A 10-gram sample of water will lose the greatest amount of heat when its temperature is changed from 50°C to
1
10°C
2
20°C
3
30°C
4
40°C
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Answer:
10°C
Explanation:
Here the problem deals with the relationship between the amount of heat lost and the temperature change of a body.
From the relationship below:
H = m C Δt
H is the quantity of heat
m is the mass
C is the specific heat
Δt is the change in heat
We can see that the quantity of heat is directly proportional to the Δt, so as more heat changes occur, so also is the value of Δt .
Related Questions
normal saline is a therapy option for severe vomiting because this solution provides _________ ions, which replace bicarbonate ions that are responsible for the metabolic imbalance.
Answers
Normal saline is a therapy option for severe vomiting because this solution provides sodium and chloride ions, which can help to replace bicarbonate ions that may be lost due to vomiting.
Bicarbonate ions play a key role in maintaining the body's acid-base balance, and their loss can lead to metabolic acidosis. By providing additional sodium and chloride ions through the administration of normal saline, the body can help to maintain its fluid and electrolyte balance, which can be disrupted during periods of vomiting.
Normal saline is a sterile solution that contains a 0.9% concentration of sodium chloride. It is often used as a replacement fluid in situations where the body has lost significant amounts of fluid and electrolytes, such as during severe vomiting or diarrhea. The sodium and chloride ions in normal saline can help to restore the body's fluid and electrolyte balance, which can be disrupted during periods of illness.
In summary, normal saline is a therapy option for severe vomiting because it provides sodium and chloride ions that can help to replace bicarbonate ions that may be lost due to vomiting. This can help to maintain the body's fluid and electrolyte balance, which is essential for proper physiological function.
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In this equation, 2mg + o2 arrow 2mgo, what is the subscript of the oxygen molecule?
Answers
The subscript of the oxygen molecule in the equation 2Mg + O2 → 2MgO is 2.
This means that two oxygen molecules (O2) are involved in the reaction, and they combine with two magnesium atoms (2Mg) to form two magnesium oxide molecules (2MgO). The subscript "2" in front of MgO indicates that two MgO molecules are formed as a result of the reaction. The equation 2Mg + O2 → 2MgO represents a combustion reaction in which magnesium (Mg) reacts with oxygen (O2) to produce magnesium oxide (MgO). This is an example of an exothermic reaction, which releases energy in the form of heat and light.
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A large fish tank is initially filled with 30 litres of fresh water. You begin to fill the tank by slowly pouring in water with salt concentration of 35 grams per litre (approximate salinity of sea water) at a rate of 2 litres per minute. At the same time, the (perfectly mixed) fluid in the tank is drained from the bottom at a rate of 1 litre per minute. 1. Determine the volume of water in the tank at time t. [1 mark] 2. Let S(t) denote the amount of salt in the fish tank at time t in grams. Show that S(t) satisfies the ODE S
′
(t)=70−
t+30
S
. Write down the appropriate initial condition for the ODE as well. [2 marks] 3. What order is this ODE? Is it linear? Is it separable? [1 mark] 4. Solve the initial value problem to find S(t) using the method of integrating factors. [3 marks] 5. What is the salt concentration in the tank as t→[infinity] ? [1 mark] Part B: Double tanks Next you hook up two fish tanks in a loop so that there is a pipe from tank A to tank B, and also a pipe from tank B back to tank A. Two pumps are added so that you can control the flow rate in each pipe. Initially tank A contains 80 litre of fresh water and tank B 60 litres of fresh water. You begin to pour salt water with concentration 35 grams per litre into tank A at a rate of 2 litres per minute. To keep the tanks from overflowing, you set your pumps so that water is flowing at a constant rate of 4 litres per minute from tank A to tank B, and 2 litre per minute from tank B to tank A. You also put a drain in tank B so that fluid is draining at a rate of 2 litres per minute. 1. Sketch a diagram of the tank setup with arrows for flows entering and leaving each tank. [
1 mark]
2. Let P(t) and Q(t) denote the amount of salt in tank A and tank B respectively. Show that P and Q satisfy a system of ODE's in the form of
P
′
(t)
Q
′
(t)
=c
1
P(t)+c
2
Q(t)+c
3
=c
4
P(t)+c
5
Q(t)
where c
1
,c
2
,c
3
,c
4
and c
5
are constants. Determine the constant c
1
,c
2
,c
3
,c
4
,c
5
and write down appropriate initial conditions. [2 marks] 3. Show that the system of ODE's can be converted into the following second order ODE for P(t) P
′′
(t)=−
60
7
P
′
(t)−
600
1
P(t)+
3
14
State the initial conditions for this ODE. [2 marks] 4. Solve this second order ODE to find P(t), and hence Q(t) as well. [4 marks ] Part C: Triple tanks You decide to have a more elaborate setup and connect three small fish tanks in a loop. You run pipes from tank C to tank D, from tank D to tank E, and finally from tank E back to tank C. Pumps are installed so that water is constantly cycling through all three pipes at the rate of 1 litre per minute. Suppose initally tank C is filled with 40 litres of salt water with a concentration 80 grams per litre, tank D with 20 litres of fresh water, and tank E 40 litres of fresh water. 1. Sketch a diagram of the tank setup with arrows for flows entering and leaving each tank. [1 mark] 2. Let X(t),Y(t),Z(t) denote the amount of salt in tanks C,D and E respectively. Write down a system of ODE's for X,Y and Z. [2 marks] 3. Rewrite your system of ODE's in the form of
⎣
⎡
X
′
(t)
Y
′
(t)
Z
′
(t)
⎦
⎤
=A
⎣
⎡
X(t)
Y(t)
Z(t)
⎦
⎤
for some matrix A. 4. Given that
⎣
⎡
2
1
2
⎦
⎤
,
⎣
⎡
−1−i
−1+i
2
⎦
⎤
, and
⎣
⎡
−1+i
−1−i
2
⎦
⎤
are eigenvectors of the matrix A, determine the corresponding eigenvalues. [2 marks] 5. Hence write down the general solution of
⎣
⎡
X(t)
Y(t)
Z(t)
⎦
⎤
. From the general solution, deduce whether the salt concentrations in the tanks settle to a final steady state as time approaches infinity, or oscillate endlessly without settling. Justify your answer.
Answers
1. The net rate at which water is being added is 2 - 1 = 1 liter per minute.
2.The initial condition for this ODE is S(0) = 0, since there is no salt in the tank initially.
3. This ODE is first-order, linear, and separable.
4. the solution to the initial value problem is \(S(t) = e^{(70t - 30ln|t|)\)
5. the salt concentration in the tank as t approaches infinity is 70 grams per liter
1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained.
The rate at which water is being added is 2 liters per minute, and the rate at which water is being drained is 1 liter per minute.
Therefore, the net rate at which water is being added is 2 - 1 = 1 liter per minute.
Since the tank initially contains 30 liters of fresh water, the volume of water in the tank at time t is given by the equation V(t) = 30 + t,
where t is the number of minutes that have passed.
2. Let S(t) denote the amount of salt in the fish tank at time t in grams. The rate at which salt is being added to the tank is 35 grams per liter multiplied by the rate at which water is being added, which is 2 liters per minute.
Therefore, the rate at which salt is being added is 35 * 2 = 70 grams per minute.
To find the derivative of S(t), we need to subtract the rate at which salt is being drained from the tank.
Since the fluid in the tank is perfectly mixed, the rate at which salt is being drained is given by the equation (1/t)(S(t) * 30),
where S(t) * 30 is the concentration of salt in the tank multiplied by the rate at which water is being drained, which is 1 liter per minute.
Therefore, we have \(S'(t) = 70 - (1/t)(S(t) * 30).\)
The initial condition for this ODE is S(0) = 0, since there is no salt in the tank initially.
3. This ODE is first-order, linear, and separable. It is first-order because it involves the derivative of S(t), linear because it is a linear combination of the function S(t) and its derivative, and separable because it can be rewritten as \(S'(t) = 70 - (30/t)S(t)\).
4. To solve the initial value problem, we can rewrite the ODE as \((1/S(t))dS(t) = (70 - (30/t))dt.\)
Integrating both sides, we get \(ln|S(t)| = 70t - 30ln|t| + C\),
where C is the constant of integration.
Exponentiating both sides, we have \(|S(t)| = e^(70t - 30ln|t| + C).\)
Since the initial condition is\(S(0) = 0\), we can substitute t = 0 into the equation and solve for C.
We get \(|0| = e^{(C)},\)
so C = 0.
Therefore, the solution to the initial value problem is \(S(t) = e^{(70t - 30ln|t|)\)
5. As t approaches infinity, the term \(e^(-30ln|t|)\) approaches 0, since the logarithm grows slower than any positive power of t.
Therefore, the salt concentration in the tank as t approaches infinity is 70 grams per liter, which is the concentration of the salt water being added to the tank.
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The volume of water in the tank at time t is 30 + t litres. The salt amount, S(t), satisfies the ODE S'(t) = 70 - t + 30/S(t) with the initial condition S(0) = 0. This is a first-order linear ODE that is not separable. The solution to the initial value problem is S(t) = (-3/2t^2 + 300t)^(1/3). As t approaches infinity, the salt concentration in the tank approaches (300t)^(1/3).
1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which it is being drained.
The rate at which water is being poured into the tank is 2 litres per minute. So after t minutes, the amount of water poured into the tank would be 2t litres.
The rate at which water is being drained from the tank is 1 litre per minute. So after t minutes, the amount of water drained from the tank would be t litres.
Therefore, the volume of water in the tank at time t can be calculated by subtracting the amount of water drained from the amount of water poured in:
Volume of water = 30 + (2t - t) = 30 + t litres.
2. To show that S(t) satisfies the ODE S'(t) = 70 - t + 30/S(t), we need to differentiate S(t) with respect to t and compare it to the right-hand side of the equation.
Differentiating S(t), we get:
S'(t) = -1 + (d/dt)(30/S(t))
= -1 + (-30/S(t)^2)(dS/dt)
= -1 - 30S'(t)/S(t)^2.
Substituting this into the original ODE, we have:
-1 - 30S'(t)/S(t)^2 = 70 - t + 30/S(t).
Simplifying the equation, we get:
-30S'(t)/S(t)^2 = 71 - t.
Multiplying both sides by -1, we have:
30S'(t)/S(t)^2 = t - 71.
Therefore, S(t) satisfies the ODE S'(t) = 70 - t + 30/S(t).
The appropriate initial condition for the ODE is S(0) = 0, as at time t = 0, there is no salt in the fish tank.
3. The order of this ODE is 1, as it involves only the first derivative of S(t). The ODE is linear, as it is in the form S'(t) = 70 - t + 30/S(t). However, it is not separable, as the variables t and S(t) are not separated on different sides of the equation.
4. To solve the initial value problem S'(t) = 70 - t + 30/S(t), with the initial condition S(0) = 0, we can use the method of integrating factors.
Multiplying both sides of the equation by S(t)^2, we get:
S(t)^2S'(t) = (70 - t)S(t)^2 + 30.
Now, let u(t) = S(t)^3. Differentiating both sides with respect to t, we have:
u'(t) = 3S(t)^2S'(t).
Substituting this into the equation, we get:
u'(t)/3 = (70 - t)S(t)^2 + 30.
Integrating both sides with respect to t, we have:
∫(u'(t)/3) dt = ∫[(70 - t)S(t)^2 + 30] dt.
Simplifying the equation, we get:
u(t)/3 = -1/2t^2 + 70t + 30t + C,
where C is the constant of integration.
Rearranging the equation, we have:
u(t)/3 = -1/2t^2 + 100t + C.
Now, substituting back u(t) = S(t)^3, we get:
S(t)^3 = -3/2t^2 + 300t + 3C.
Taking the cube root of both sides, we have:
S(t) = (-3/2t^2 + 300t + 3C)^(1/3).
By applying the initial condition S(0) = 0, we can solve for the constant C:
0 = (-3/2(0)^2 + 300(0) + 3C)^(1/3),
0 = 3C,
C = 0.
Therefore, the solution to the initial value problem is:
S(t) = (-3/2t^2 + 300t)^(1/3).
5. As t approaches infinity, the term -3/2t^2 becomes negligible compared to 300t. Thus, the salt concentration in the tank as t approaches infinity is approximately given by S(t) = (300t)^(1/3).
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Lexi is researching crime rates. She is trying to make a conclusion about whether the crime rate changes with the population in each county in her state. What is the outcome variable (dependent variable) of her investigation?
Answers
Lexi is researching crime rates. The variable being measured or tested in an experiment is known as the dependent variable. The results of the participants' tests, for instance, since that is what is being measured, would be the dependent variable in a study looking at how tutoring affects test scores.
What is dependent variable ?In experimental sciences, there are dependent and independent variables. Dependent variables get their name because, during an experiment, their values are investigated under the presumption or requirement that they are dependent on the values of other variables due to some law or rule.
By dividing the total number of reported offenses by the population, the crime rate is determined. After then, the result is multiplied by 100,000. In 2014, for instance, there were 48,650 robberies committed in California, which had a population of 38,499,378. This results in a crime rate for robberies of 126.4 per 100,000.
Thus, the dependent variable in a study looking at how tutoring affects test scores.
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Propane is widely used in liquid form as a fuel for barbecue grills and camp stoves. For 85.5 g of propane, calculate
(b) grams of carbon.
Answers
Propane is widely used in liquid form as a fuel for barbecue grills and camp stoves. For 85.5 g of propane, grams of carbon is 3.55 x 10²⁴ molecules.
What is the calculations?Given:
moles of propane = 5.9 moles
molecules of propane (C₈H₈) = ?
Formula: no. of moles = no. of molecules / Avogadro's number
We use:
no. of molecules = no. of moles x Avogadro's number . . . . . (1)
Where,
Avogadro's number = 6.022 x 10²³
Put values in equation (1)
no. of molecules =5.9 mol x 6.022 x 10²³ (molecules/mol)
no. of molecules = 3.55 x 10²⁴ molecules
So, There are 3.55 x 10²⁴ molecules in 5.9 moles of propane
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This water bottle has a mass of 0.5 kg.
CU
If the net force acting on the bottle is 5 N to the right, what is the bottle's
acceleration?
(Hint: Use the formula a = =)
A. a = 10 m/s2 left
B. a = 50 m/s2 right
c. a = 10 m/s2 right
D. a = 2 m/s2 left
PLEASE ANSWER I WILL GIVE BRAINLYEST
Answers
Answer: the answer is a=10m/s^2 right
Explanation: it just make more since then the other answers .
Answer:
c 10
Explanation:
ape.x verified
i need help solving this
Answers
3) 1.05 is the correct answer
Explanation:9.7 - 8.65 = 1.05
Answer:
1.05
Explanation:
8.65-9.7=-1.05cm
what is the molar mass of a covalent compound if 0.995 g of it is dissolved in 24 ml of water and produces a freezing temperature of -0.64oc?
Answers
77.4 g/mol is the molar mass of a covalent compound if 0.995 g of it is dissolved in 24 ml of water and produces a freezing temperature of -0.64°C
To solve for the molar mass of the covalent compound, we need to use the freezing point depression equation:
ΔT = Kf m i
where ΔT is the change in freezing point (in Celsius), Kf is the freezing point depression constant for water (1.86 °C/m), m is the molality of the solution (in moles of solute per kilogram of solvent), and i is the van't Hoff factor (which is 1 for a covalent compound).
First, we need to calculate the molality of the solution:
m = moles of solute / mass of solvent (in kg)
Since 0.995 g of the compound is dissolved in 24 ml of water, the mass of solvent is 24 g (since the density of water is 1 g/mL). Therefore,
m = moles of solute / 0.024 kg
moles of solute = (0.995 g / molar mass of compound) / 0.024 kg
Next, we can plug in the given values for ΔT (-0.64 °C), Kf (1.86 °C/m), and m (calculated above) to solve for the molar mass of the compound:
-0.64 = 1.86 × [(0.995 / m) / 0.024]
molar mass of compound = 77.4 g/mol
Therefore, the molar mass of the covalent compound is 77.4 g/mol.
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please help me
16 1 point What is the decay rate of a sample of Oxygen-21 if the sample has 8.31x1017 atoms and a decay constant of 0.203/s? 4.09x1018Bq 1.69x10¹7Bq 0.203Bq 2.44x10-1⁹Bq Previous
Answers
decay rate of approximately 1.69x10^17 Bq (becquerels),
The decay rate of a radioactive sample is determined by the number of radioactive atoms present and the decay constant, which represents the probability of decay per unit of time.
To calculate the decay rate, we multiply the number of atoms in the sample by the decay constant. In this case, the sample has 8.31x10^17 atoms and a decay constant of 0.203/s. Multiplying these values gives a decay rate of approximately 1.69x10^17 Bq (becquerels), which represents the number of decays per second in the sample.
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Which General Hazardous Materials Behavior Model event involves the material being free to travel or disperse, allowing it to move outward from the source?
Answers
The General Hazardous Materials Behavior Model event that involves the material being free to travel or disperse, allowing it to move outward from the source is known as dispersion.
This is the third event in the model and occurs after the release and contact events. During dispersion, the material is no longer contained and can spread through the air, water, or ground.
This can potentially cause harm to people, animals, and the environment. It is important to understand and predict the dispersion of hazardous materials in order to prevent or minimize the impact of an incident.
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what is the balanced equation of magnesium carbonate decomposes on heating to form magnesium oxide and carbon dioxide
Answers
Answer:
ExplaAt high temperatures MgCO3 decomposes to magnesium oxide and carbon dioxide. This process is important in the production of magnesium oxide. This process is called calcining: MgCO3 → MgO + CO2 (ΔH = +118 kJ/mol)
nation:
How has the way humans communicate changed over time?
Answers
What are some possible factors that must remain constant during the testing
Answers
Answer:
Four basic components that affect the validity of an experiment are the control, independent and dependent variables, and constants. These basic requirements need to be present and identified to consider an experiment valid.
When we have stomach disorder, we take antacid to solve the problem. How does it help? Support your answer with an equation.
Answers
Answer:
its help us
equation- ( a)2+(b)2+2an
What is the chemical formula for aloe vera gel (or just aloe vera)?
Answers
Answer:
Aloeride is a polysaccharide that comprises only 0.015% of the crude A. vera juice material (dry weight). It has a molecular weight between 4 and 7 million Da with its glycosyl components containing glucose (37.2%), galactose (23.9%), mannose (19.5%) and arabinose (10.3%).
Answer:
Aloeride is a polysaccharide that comprises only 0.015% of the crude A. vera juice material (dry weight). It has a molecular weight between 4 and 7 million Da with its glycosyl components containing glucose (37.2%), galactose (23.9%), mannose (19.5%) and arabinose (10.3%).
Explanation:
in radioactive decay, which type of emission involves an unstable nucleus creating an electron then ejecting it in order to become stable?
Answers
As radioactive atoms decay and try to come to be stable, the nuclei launch strength withinside the shape of ionizing radiation (alpha debris, beta debris and gamma rays).
The strength launched is known as ionizing radiation as it has sufficient strength to knock tightly certain electrons from the atom's orbit. Alpha decay happens whilst the nucleus ejects an alpha particle (helium nucleus). Beta decay happens in ways; beta-minus decay, whilst the nucleus emits an electron and an antineutrino in a method that adjustments a neutron to a proton. Radioactive decay — Disintegration of the nucleus of an volatile atom with the aid of using the discharge of radiation. Radioactivity — The method of spontaneous transformation of the nucleus, normally with the emission of alpha or beta debris frequently followed with the aid of using gamma rays.
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Draw a dash-wedge structure for (2S,3R)-3-bromo-6,6-dimethylocta-7-en-2-ol. Draw a dash-wedge structure for (3S,4R)-4-chloro-3,5-dimethylhex-1-yne.
Answers
This is the dash-wedge structure for (2S,3R)-3-bromo-6,6-dimethylocta-7-en-2-ol:
Br
|
H3C–C–CH=CH–CH2–C(CH3)2–OH
| | | |
CH3 CH3 CH3 H
wedge
This is the dash-wedge structure for (3S,4R)-4-chloro-3,5-dimethylhex-1-yne:
Cl
|
CH3–C≡C–CH(CH3)–CH(CH3)2
| | |
H CH3 CH3
wedge dash
The dash-wedge structure is a way of representing three-dimensional molecular structures on a two-dimensional surface. In this notation, solid lines represent bonds that are in the plane of the paper or screen, dashed lines represent bonds that are going away from the viewer (into the paper or screen), and wedge-shaped lines represent bonds that are coming out of the viewer (towards the viewer). This notation helps us to visualize the spatial arrangement of atoms in a molecule, which is important for understanding the molecule's properties, reactivity, and interactions with other molecules.
In the first molecule, (2S,3R)-3-bromo-6,6-dimethylocta-7-en-2-ol, the stereochemistry is specified by the two stereocenters at positions 2 and 3. The S and R designations refer to the absolute configuration of the stereocenters, determined by the Cahn-Ingold-Prelog (CIP) priority rules. The bromine atom is attached to the stereocenter at position 3, and its orientation is shown with a wedge-shaped bond, indicating that it is coming out of the page towards the viewer. The hydroxyl group at position 2 is shown with a dashed bond, indicating that it is going away from the viewer, and the other atoms are shown with solid lines. The methyl groups on positions 6 and 8 are both in the plane of the paper, and the other methyl group at position 7 is going away from the viewer.
In the second molecule, (3S,4R)-4-chloro-3,5-dimethylhex-1-yne, there is only one stereocenter at position 3, which has an S configuration, and the other stereodescriptor, R, refers to the chirality at position 4. The triple bond between carbons 1 and 2 is shown with a straight line, and the chlorine atom at position 4 is shown with a wedge-shaped bond, indicating that it is coming out of the page towards the viewer. The two methyl groups at positions 3 and 5 are both in the plane of the paper, and the other methyl group at position 6 is shown with a dashed bond, indicating that it is going away from the viewer. The hydrogen atom at position 1 is also going away from the viewer, and the other hydrogen atoms are not shown for clarity.
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How is the gravitational force related to the distance between two objects?
O It is directly proportional to the square of the distance.
O It is inversely proportional to the square of the distance.
OIt is directly proportional to the distance.
O It is inversely proportional to the distance.
Answers
The gravitational force is related to the distance between two objects because it is inversely proportional to the square of the distance.
What is the gravitational force?The gravitational force refers to the attraction between two different objects due to their masses.
This force (gravitational force) is fundamental for maintaining the distance of objects in the Universe.
In conclusion, the gravitational force is related to the distance between two objects because it is inversely proportional to the square of the distance.
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The number and diversity of plants are different in different biomes and is influenced mostly by which factors?
OPTIONS
The number of grazing animals and insects
The strength and direction of winds
The amount of sunlight and precipitation
Seasonal changes in temperature and humidity
Answers
The amount of energy from the sun and precipitation level are critical factors in determining the type of vegetation in each terrestrial biome. These biomes include grassland, desert, tundra and forest.
The number and diversity of plants are different in different biomes because they are mostly influenced by the amount of sunlight and precipitation.In grassland, the dominant vegetation is grasses. In this biome, the precipitation level is moderate. There are two types of grassland depending on the latitude (i.e., depending on the amount of sunlight): savanna and temperate grasslands. In desert biomes, the precipitation level is low and the sunlight is high. These biomes are characterized by annual plants (generally annual grasses).In forests, the precipitation level is high and the amount of sunlight is high. This biome is characterized by trees as dominant vegetation.In the tundra, the precipitation level is low and the amount of sunlight is low. This biome is characterized by short-growing plants.Learn more in:
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1. Explain the differences between the "lonic Bond" and "Covalent Bond". 2. 500 grams of sugar occupies a volume of 0.315 L. What is the density of the sugar in g/cm^3 ? 3. What is the mass of a 450 m^3 block of silicon if the density of silicon is 2.336 g/cm^3 ?
Answers
1. The ionic bond and covalent bond are distinguished from each other based on the way the atoms are attached to each other. Ionic bonds are formed when one or more electrons are transferred from one atom to another, resulting in a cation and an anion attracting each other to form an ionic bond. Ionic bonds are a result of electrostatic attraction between ions with opposite charges. Covalent bonds are formed when atoms share one or more pairs of electrons to achieve stable electronic configurations. Covalent bonds form when two or more atoms share electrons in order to achieve the stable electron configuration of noble gases. In covalent bonding, the electrons are shared between atoms, not transferred.
2. Density is defined as mass per unit volume. The density of sugar in g/cm³ is obtained by dividing its mass by its volume.
Mass of sugar = 500gVolume of sugar = 0.315L
Using the formula for density, we have
Density = Mass/Volume= 500g/0.315
L= 1587.30 g/L1
L = 1000 mL1 mL = 1 cm³
Density = 1587.30 g/L × (1 mL/1 cm³)
Density = 1.5873 g/cm³, to 4 significant figures
3. We can use the formula; Density = Mass/Volume, to calculate the mass of the block of silicon. Volume of block of silicon = 450 m³ Density of silicon = 2.336 g/cm³ Volume of block of silicon = 450 m³ = 4.5 × 10^8 cm³
We can substitute the values into the formula
Density = Mass/Volume
2.336 g/cm³ = Mass/4.5 × 10^8 cm³
Mass = 2.336 g/cm³ × 4.5 × 10^8 cm³
Mass = 1051200000 g or 1.05 × 10^9 g, to 2 significant figures.
Therefore, the mass of the block of silicon is 1.05 × 10^9 g.
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which of the following is unbalanced
A. SnO2 + H2 → Sn + 2 H2O
B. 2Na + 2H2O → 2NaOH + H2
C. N2O5 + H2O → HNO3
D. 2Fe + 3Cl2 → 2FeCl3
Answers
Answer:
The unbalanced chemical equation is;
C. N₂O₅ + H₂O → HNO₃
Explanation:
In balancing a chemical reaction equation, only the coefficients of the terms of the reactants and/or the products are adjusted (and not the subscripts to the atoms in the compounds) to ensure that the total number of each element is the same on either side of the arrow for the chemical reaction
A coefficient term is the number multiplier of the compounds and elements in a chemical reaction
Therefore, the balanced chemical equation is obtained by changing the coefficient of the product from the implied '1' to a '2' as follows;
N₂O₅ + H₂O → 2HNO₃
suppose you have two samples that are equal in weight, 22.3 g zn and 22.3 g cr2o3 . calculate the number of moles of each substance.
Answers
The Number of Moles of Zn is 0.34108 and the number of moles of Cr2O3 is 0.14671.
What is Number of Moles?
The ratio of a substance's given mass in a chemical reaction to the mass of one mole of that material is what determines how many moles there are of that substance. A mole of any substance is equal to 6.0231023, or Avogadro's number. The ratio of a substance's given mass in any kind of chemical reaction to the mass of one mole of that material is the number of moles of that substance.
Moles of Zn can be calculated by using the formula
22.3 g of Zn x 1 mole Zn/65.4 g = 0.341 moles of Zn.
Moles of Cr2O3 can be calculated by using the formula
22.3 g of Cr2O3 x 1 mole Cr2O3/151.99 = 0.146 moles of Cr2O3.
Hence, 0.34108 and 0.14671 will be the number of moles for Zn adn Cr2O3 respectively.
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Zn has a mole count of 0.34108, while Cr2O3 has a mole count of 0.14671.
What is Number of Moles?
The number of moles of a substance is determined by the ratio of its given mass in a chemical reaction towards the mass of one mole of that substance. A mole, or 6.0231023, is equal to one mole of any substance. The number of moles of a substance is defined as the mass of one mole divided by the given mass of the substance in any type of chemical reaction.
The formula can be used to determine moles of zinc.
22.3 g of Zn x 1 mole Zn/65.4 g = 0.341 moles of Zn.
The formula can be used to determine moles of Cr2O3.
22.3 g of Cr2O3 x 1 mole Cr2O3/151.99 = 0.146 moles of Cr2O3.
Therefore, the moles of Zn and Cr2O3 will be 0.34108 and 0.14671, respectively.
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which of the following is a compound .a)oxygen b) Magnesium c) Carbon dioxide d) Argon
Answers
C)Carbondixoide is a compound as it contains both carbon and oxygen.
A)oxygen is a element with symbol O
B)Magnesium is a element with symbol Mg
C> carbon dioxide is CO2
D)Argon is Ar
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a 15.0 ml solution of ba(oh)₂ is neutralized with 22.7 ml of 0.200 m hcl. what is the concentration of the original ba(oh)₂ solution?
Answers
The concentration of the original Ba(OH)₂ solution if 15.0 ml solution of Ba(OH)₂ is neutralized with 22.7 ml of 0.200 m HCl is 151.3 mol/dm³
To determine concentration of the original Ba(OH)₂ solution, we must know he balanced chemical equation for the neutralization reaction is:
Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O
From the equation above, the stoichiometric ratio of Ba(OH)₂ and HCl is 1:2. That means one mole of Ba(OH)₂ reacts with 2 moles of HCl. The balanced chemical equation also shows that the number of moles of HCl used is the same as the number of moles of Ba(OH)₂. Hence:
moles of HCl = 0.200 mol/dm³ × 22.7 dm³ = 4.54 mol
Using the stoichiometric ratio, the moles of Ba(OH)₂ in the solution can be calculated to be:
moles of Ba(OH)₂ = 4.54 mol ÷ 2 = 2.27 mol
The volume of the Ba(OH)₂ solution is 15.0 mL, which is 0.015 dm³. Therefore, the concentration of the original Ba(OH)₂ solution can be calculated as:
concentration = moles/volume= 2.27 mol ÷ 0.015 dm³= 151.3 mol/dm³
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The concentration of the original Ba(OH)₂ solution is 0.302 M.
Given data
Volume of Ba(OH)₂ solution used = 15.0 ml
Volume of HCl used = 22.7 ml
Molarity of HCl solution used = 0.200 M
We need to calculate the concentration of Ba(OH)₂ solution, which is not known.Molar ratio of HCl and Ba(OH)₂ in a balanced chemical equation of their neutralization is;
HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
The balanced chemical equation tells us that 1 mole of HCl is required to neutralize 1 mole of Ba(OH)₂.
So, the moles of HCl used in the reaction is;
moles of HCl = molarity × volume (in liters)
moles of HCl = 0.200 M × 0.0227 L = 0.00454 mole
Since one mole of HCl reacts with 1 mole of Ba(OH)₂,
so the number of moles of Ba(OH)₂ used is also equal to 0.00454 mole. Since we know the volume of the Ba(OH)₂ solution used, we can calculate the molarity of the solution as;
molarity = moles of solute / volume of solution in liters
Molarity = 0.00454 / (15.0 / 1000) = 0.302 M
Therefore, the concentration of the original Ba(OH)₂ solution is 0.302 M.
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No underwater luminaires shall be installed in a permanently-installed swimming pool that operates on supply circuits over ? between conductors.
Answers
No luminaires shall be installed for operation on supply circuits over 150 volts between conductors. Luminaires mounted in walls shall be installed with the top of the luminaire lens not less than 450 mm (18 in.)
In an electrical installation, what is a luminaire?
A luminaire, also known as a light fixture, is a complete lighting assembly that includes one or more lamps (light bulbs or tubes), a socket and other protective elements, wiring that links the lamp to a power source, and a reflector to help direct and spread the light.
The construction, installation, or inclusion of shades or guards on luminaires is required to prevent the exposure of combustible materials to temperatures above 90°C (194°F). Unswitched lamp holders must be used above materials that are very flammable.
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what atoms are present In the molecule KNO3 and how many are there of each ?
Answers
Answer:
one potassium, one nitrogen and three oxygen
Explanation:
potassium is K, nitrogen is N and oxygen is O
if carbon dioxide is broken down. what element will it give?
Answers
Identify each of the following as a fatty acid, soap, triacylglycerol, wax, glycerophospholipid, sphingolipid, or steroid.
Part B
whale blubber
fatty acid
soap
triacylglycerol
wax
glycerophospholipid
sphingolipid
steroid
Part C
tristearin
fatty acid
soap
triacylglycerol
wax
glycerophospholipid
sphingolipid
steroid
Part D
progesterone
fatty acid
soap
triacylglycerol
wax
glycerophospholipid
sphingolipid
steroid
Part E
cortisone
fatty acid
soap
triacylglycerol
wax
glycerophospholipid
sphingolipid
steroid
Part F
stearic acid
fatty acid
soap
triacylglycerol
wax
glycerophospholipid
sphingolipid
steroid
Answers
The classification for each item: Part B - Whale blubber: triacylglycerol Part C - Tristearin: triacylglycerol Part D - Progesterone: steroid Part E - Cortisone: steroid Part F - Stearic acid: fatty acid
Part B:
- Whale blubber: triacylglycerol
- Fatty acid: fatty acid
- Soap: soap
- Wax: wax
- Glycerophospholipid: glycerophospholipid
- Sphingolipid: sphingolipid
- Steroid: steroid
Part C:
- Tristearin: triacylglycerol
- Fatty acid: fatty acid
- Soap: soap
- Wax: wax
- Glycerophospholipid: glycerophospholipid
- Sphingolipid: sphingolipid
- Steroid: steroid
Part D:
- Progesterone: steroid
- Fatty acid: fatty acid
- Soap: soap
- Wax: wax
- Glycerophospholipid: glycerophospholipid
- Sphingolipid: sphingolipid
- Steroid: steroid
Part E:
- Cortisone: steroid
- Fatty acid: fatty acid
- Soap: soap
- Wax: wax
- Glycerophospholipid: glycerophospholipid
- Sphingolipid: sphingolipid
- Steroid: steroid
Part F:
- Stearic acid: fatty acid
- Soap: soap
- Triacylglycerol: triacylglycerol
- Wax: wax
- Glycerophospholipid: glycerophospholipid
- Sphingolipid: sphingolipid
- Steroid: steroid
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What is the mass of 3.00x10^23 particles of water? Water has a molar mass of 18.0 g/mol.
Answers
8.97 g H₂O
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Atomic Structure
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.Stoichiometry
Using Dimensional AnalysisExplanation:Step 1: Define
3.00 × 10²³ particles H₂O
Step 2: Identify Conversions
Avogadro's Number
[Given] Molar Mass of H₂O - 18.0 g/mol
Step 3: Convert
Set up: \(\displaystyle 3.00 \cdot 10^{23} \ particles \ H_2O(\frac{1 \ mol \ H_2O}{6.022 \cdot 10^{23} \ particles \ H_2O})(\frac{18.0 \ g \ H_2O}{1 \ mol \ H_2O})\)Multiply/Divide: \(\displaystyle 8.96712 \ g \ H_2O\)Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
8.96712 g H₂O ≈ 8.97 g H₂O
5.Study the reaction.2HCl (g)+I2 (s)⇌2HI (g)+Cl2 (g)When the reaction is at equilibrium, some amount of HI is added. The system then reaches equilibrium again.How does the concentration of Cl2 change compared to the first equilibriumThe concentration is zero.The concentration is higher.The concentration is lower.The concentration is the same.
Answers
The concentration of Cl2 changes compared to the first equilibrium in the reaction 2HCl (g)+I2 (s)⇌2HI (g)+Cl2 (g) when some amount of HI is added is that the concentration of Cl2 is lower. This is because the addition of HI shifts the equilibrium to the left, causing the concentration of Cl2 to decrease. This is an example of Le Chatelier's principle, which states that when a stress is applied to a system at equilibrium, the system will shift in the direction that reduces the stress. In this case, the stress is the addition of HI, and the system shifts to the left to reduce the concentration of HI, resulting in a lower concentration of Cl2. Therefore, the correct answer is option C, "The concentration is lower."
Equilibrium is defined as a condition where the interactions that occur between the components that exist in human life activities can run in harmony and balance, and have a significant impact on human welfare.
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