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2KCIO3 → 2KCI + 302
How many grams of potassium chloride are produced if 52 g of potassium chlorate decompose?

A )32 g KCI
B )19 g KCI
C)44 g KCI
D)1.5 g KCI

Answer:

The wolf gets energy from other Animals through Cellular respiration. it's a carnivore

The rabbit gets energy from Carbohydrates,Fats.... obtained through different sources. A common example is the grass. It's an herbivore

The plant gets energy from the sun during photosynthesis. It's Autotrophic.

The mushroom gets energy from the decomposition of other organic matter. It's heterotrophic.

Explanation:

In a food chain; The Wolf eats the rabbit, when the Wolf dies, decomposers such as mushrooms breaks down its body returning it to the soil, where it provides nutrients for plants

1. Let's write the silver chromate using the oxidation number of each compound.

Oxygen has a -2 oxidation number, silver only contains +1 and chromate contains +2, 3, 6.

Remember that the ion chromate contains:

So, in our case we're going to use 2 atoms of Ag to complete the formula, like this:

2. Calculating the positive ions, in this case, corresponds to silver and chromate. In the last question, we realize that the ions of silver are +2 and the chromium was 6+. The oxidation number of oxygen is always negative so, this doesn't count. The positive ions are 2*2+6*1= +10 ions.

3. As we see before, oxygen is the only one that contains negative ions, so, we have 4 moles of oxygen which you can see in the formula. Multiplying this number by -2, we obtain -8 ions.

4. To calculate the mass we're going to use the molar mass of each element where you can find on the periodic table. (molar mass of oxygen = 16 g/mol, chromium = 52 g/mol and silver = 107.8 g/mol).

We multiply each molar mass with each mole of each element and then we're going to sum all of them, like this:

The mass of a chemical formula unit is 331.6 g/mol.

Answer:

A.Photosynthesis

B.Glucose

C . fermentation

D.Aerobic respiration

Explanation:

A. glucose form as a product of photosynthesis

B.glucose is simple sugar and it must be always converted to complex sugar so that it can be stored

C.Energy converted to ATP

D.Living organisms respire using oxygen and carbon dioxide is always formed as a by product

Answer:

the awnser is c

Explanation:

Answer:

yes

Explanation:

Answer:

yes

Explanation:

Humans inhale oxygen and breath out carbon dioxide.

The mass of P-32 remaining after 27 days, if the initial mass is 40g is 10.5 g.

The half-life of a radioactive isotope is the amount of time it takes for one-half of the radioactive isotope to decay.

The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.

The number of half lives in 27 days;

n = 27 days/14 days

n = 1.929

The mass of P-32 remaining after 27 days, if the initial mass is 40g;

mass remaining = 40 g / (2^1.929)

mass remaining = 40 g / 3.808

mass remaining = 10.5 g

Thus, the mass of P-32 remaining after 27 days, if the initial mass is 40g is 10.5 g.

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The amount of dry solute it would take to prepare 118 mL of 0.105 M NaNO3 is 1.05g.

The mass of a substance can be calculated by multiplying the number of moles in the substance by its molar mass as follows:

mass = no of moles × molar mass

However, the number of moles it took to prepare 118 mL of 0.105 M NaNO3 must be calculated as follows;

no of moles = 0.105 × 0.118 = 0.0124 moles

molar mass of sodium nitrate = 85g/mol

mass = 85g/mol × 0.0124 moles = 1.05g

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Answer: The quantitative interpretation of the reaction is as follows: 84.01 g (1 mol) of sodium

bicarbonate reacts with 36.46 g (1 mol) of hydrochloric acid to generate 58.44 g (1 mol) of salt,

18.02 g (1 mol) of water, and 44.01 g (or 1 mol) of carbon dioxide. Of course the starting

quantity of NaHCO3 may be more or less than 84.01 g, but a proportionate quantity of the

hydrochloric acid will be consumed, and proportionate quantities of the products will be formed.

For example, should 100.00 g of NaHCO3 react with excess acid, the theoretical mass of salt

produced would be calculated as follows:

58.44 g sodium chloride 100.00 g sodium bicarbonate

84.01 g sodium bicarbonate = 69.55 g sodium

chloride

Note that when a bicarbonate is reacted with excess acid, the salt produced is the only substance

not readily volatile. That is, the unreacted acid, the water, and the carbon dioxide are easily

removed by heating. Does this help?

Answer:

C.) No. of electrons

Explanation:

A.) is incorrect. The atomic number represents the number of protons in an element. Nitrogen (N) and sodium (Na) always have a differing amount of protons.

B.) is incorrect. The mass number represents the number of protons and neutrons in an element. The number of neutrons and protons are specific to each element (disregarding isotopes). When elements ionize, these amounts are not altered.

C.) is correct. When an element becomes an ion, the number of electrons change. When nitrogen gains 3 electrons and sodium loses 1 electron, they end up having the same number of electrons (10).

D.) is incorrect. When elements ionize, the number of neutrons does not change. The only way two different elements could have the same number of neutrons is if at least one of the elements is an isotope. Isotopes are two or more atoms of the same element that differ in their amounts of neutrons.

Approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M.

To determine the mass of glucose (C6H12O6) required to prepare a 0.215 M solution in 5000 mL, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's convert the volume of the solution from milliliters (mL) to liters (L):

5000 mL = 5000/1000 = 5 L

Now, we can rearrange the formula to solve for moles of solute:

moles of solute = Molarity (M) x volume of solution (L)

moles of solute = 0.215 M x 5 Lmoles of solute = 1.075 mol

Since glucose (C6H12O6) has a molar mass of approximately 180.16 g/mol, we can calculate the mass of glucose using the equation:

mass of solute = moles of solute x molar mass of solute

mass of glucose = 1.075 mol x 180.16 g/mol

mass of glucose = 194.0 g (rounded to three significant figures)

Therefore, approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M. It's important to note that the molar mass of glucose used in this calculation may vary slightly depending on the level of precision required.

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Answer:

Example 1

palladium(II) nitrate,

Example 2

Metal catalysts such as Fe, Ni, Mo, and Co are routinely used in the manufacture of CNMs.

Explanation

The three metals used in catalytic converters — rhodium, platinum and palladium — are part of a category known as platinum group metals, or PGMs, which are known for their catalytic properties.

Answer:

B. HOPE THIS HELPED SENSI

Explanation:

Answer:

2 atm

Explanation:

The symbol 'atm' is used for pressure. Hence, the symbol used to indicate the pressure at which a chemical reaction is carried out can be, 2 atm.

Answer:

vinegar and soap

Explanation:

Answer:

vinegar and soap

Explanation: just took the test

Answer: The final pressure is 75 mm Hg.

Explanation:

According to Boyle's law, at constant temperature the pressure of a gas in inversely proportional to volume.

Since, it is given that the temperature is constant. Hence, formula used is as follows.

\(P_{1}V_{1} = P_{2}V_{2}\)

Substitute the values into above formula as follows.

\(P_{1}V_{1} = P_{2}V_{2}\\750 mm Hg \times 2 L = P_{2} \times 20 L\\P_{2} = \frac{750 mm Hg \times 2 L}{20 L}\\= 75 mm Hg\)

Thus, we can conclude that the final pressure is 75 mm Hg.

A) The specific humidity of the air with the given parameters is;

w1 = 0.00967

B) The relative humidity of the air with the given parameters is;

Φ1 = 0.459

C) The enthalpy of the air in KJ/kg dry air with the given parameters is;

h1 = 49.75 KJ/Kg

Correct question is;

The dry- and wet-bulb temperatures of atmospheric air at 95 kPa are 25 and 17 °C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air, in kJ/kg dry air.

We are given;

Atmospheric Pressure;P = 95 KPa

Dry temperature;T1 = 25 °C

Wet temperature;T2 = 17°C

A) From table A-4 attached and at temperature of 17°C and by interpolation, we have a saturation pressure of P_g2 = 1.938 kpa

First of all, we will calculate the specific humidity from the given pressure and saturation pressure with the formula;

w2 = (0.622 × P_g2)/(P - P_g2)

w2 = (0.622 × 1.938)/(95 - 1.938)

w2 = 0.013

Now, let's calculate specific humidity with the enthalpies at 17 °C and by interpolation. We have specific enthalpies from table A-4 as;

h_fg2 = 2460 KJ/Kg

h_g1 = 2546.5 KJ/Kg

h_f2 = 71.36 KJ/Kg

The formula for the specific humidity under these conditions is;

w1 = (c_p(T2 - T1) + w2•h_fg2)/(h_g1 - h_f2)

c_p of air has a value of 1.005 KJ/Kg.°C

Thus;

w1 = (1.005(17 - 25) + 0.013*2460)/(2546.5 - 71.36)

w1 = 0.00967

B) The relative humidity is determined from the equation;

Φ1 = (w1*p)/(0.622 + w1)p_g1

From table A-4 attached and at temperature of 25 °C, we have a saturation pressure of P_g1 = 3.1698 KPa

Φ1 = (0.00967*95)/(0.622 + 0.00967)3.1698

Φ1 = 0.459

C) For the enthalpy of air, h1 we will use the formula;

h1 = (c_p × T1) + (w1 × h_g1)

h1 = (1.005 × 25) + (0.00967 × 2546.5)

h1 = 49.75 KJ/Kg

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Answer:

A water molecule is made up of two hydrogen atoms and one oxygen atom. A single oxygen atom contains six electrons in its outer shell, which can hold a total of eight electrons. When two hydrogen atoms are bound to an oxygen atom, the outer electron shell of oxygen is filled.

Explanation:

Water has millions of molecules of liquid....Clear liquid, So that's how I came up with my answer...

Because it's the molecule.

And the way it's said is H2O.

The combined gas law equation is:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

The volume of the gas at 30 °C is approximately 760.67 mL.

To determine the volume of the gas at 30 °C, we can use the combined gas law equation, which relates the initial and final conditions of temperature and volume for a gas.

The combined gas law equation is:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:

P1 and P2 are the initial and final pressures, respectively

V1 and V2 are the initial and final volumes, respectively

T1 and T2 are the initial and final temperatures in Kelvin, respectively

We need to convert the temperatures from Celsius to Kelvin by adding 273.15 to each value.

Given:

V1 = 550 mL

T1 = -55 °C = 218.15 K

T2 = 30 °C = 303.15 K

Assuming the pressure remains constant, we can rearrange the equation to solve for V2:

V2 = (P1 * V1 * T2) / (P2 * T1)

Since the pressure is not specified in the problem, we can assume it remains constant, allowing us to cancel out the pressure terms. Thus, the final equation becomes:

V2 = (V1 * T2) / T1

Plugging in the given values:

V2 = (550 mL * 303.15 K) / 218.15 K

Simplifying the calculation, we find:

V2 ≈ 760.67 mL

Therefore, the volume of the gas at 30 °C is approximately 760.67 mL.

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The equation for the equilibrium constant (K) of the reaction studied in this exercise can be written as follows: K = ([\(CI_20\), 2-] * [\(H_20\)(1)]) / ([\(C0_4^ 2\)-] * [Ht])

In this equation, the concentrations of the species involved in the reaction are represented by the square brackets [ ]. The subscripts indicate the stoichiometric coefficients of each species in the balanced chemical equation.

The reaction being studied involves the following species:

\(C0_4^ 2\)- (ag) + 2Ht (ag) = \(CI_20\), 2- (ag) + \(H_20\)(1)

In the equilibrium constant expression, the concentration of \(CI_20\), 2- is multiplied by the concentration of \(H_20\)(1) and divided by the product of the concentrations of \(C0_4^ 2\)- and Ht. The stoichiometric coefficients in the balanced equation are used as exponents for the concentrations of the respective species.

It is important to note that the concentrations used in the equilibrium constant expression should be in molar units (mol/L) or expressed as partial pressures for gases.

Additionally, the equilibrium constant is specific to a given temperature, and its value provides information about the relative amounts of reactants and products at equilibrium.

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Step 1 - Understanding the types of heat transfer

There are three types of heat transfer: conduction, convection and irradiation. Let's see how each one works:

a) Conduction: is when a very hot substance enters in contact with a cooler substance. The atoms in the hotter substance are moving with a greater velocity and will therefore collide with the atoms of the cooler substance.

b) Convection: it happens especially in liquids and air (fluids). The liquid is warmed up at the bottom first, via conduction. The heated bottom thus changes its density: it becomes less dense, and go up. This process repeats itself several times, warming the whole liquid.

c) Irradiation: when something is put near a heat source, but do not touch it directly. In this case, there's no direct atom collision, but indirect: the rapid atoms of the heat source collide with atoms in the air which then collide with the atoms in the object that is being heated.

Step 2 - Discovering the type of heat transfer in each scenario

In the first scenario, the water at the bottom is in direct contact with the pan. Therefore, water molecules will directly collide with "pan" molecules (probably aluminum atoms or other materials). But there's also convection. The water is not heated only by conduction.

In the second scenario, the hands are near the Bunsen burner, but not directly touching it. What is happening here is then heat irradiation, not conduction.

In the third scenario, the hand is touching the object (the handle of the pan). Even though handles are made of material that poorly conduct heat, it will increase its temperature, at least a little bit. When we touch it, we can feel it is hotter than before. In this case, only conduction is involved (from the handle to our hand). In this case atoms are directly colliding and this is the only source of heat.

Finally, in the last scenario, we also have a case of irradiation: the pan with water is near the heat source, but not directly touching it.

Step 3 - How to set an experiment

In science in general, anytime we want to investigate some effect it is good manners to investigate a system controling all other effects. We want to investigate a variable at a time.

Therefore, while both scenario 1 and 3 involve conduction, scenario 1 also involves convection, which could be a problem to a experiment intending to study conduction only.

The best experiment would be then scenario 3, hands touchind the handle of pan siting in a Bunsen burner.

Answer:

.Heat energy can be transferred from one object to another. The transfer or flow due to the difference in temperature between the two objects is called heat.

Explanation:

The transfer of thermal energy is called heat. In this process, thermal energy moves through a substance.  For example, an ice cube has heat energy and so does a glass of lemonade.

Answer:

$474.30

Explanation:

23.25 × 102 = 2371.5 (total amount of money produced from pallets)

2371.5/5 = 474.3 (payout per worker)

Answer:

23 atoms for the first question

Explanation:

To solve such this we must know the concept of combination reaction. Therefore, the given reaction CH\(_4\)+Br\(_2\) \(\rightarrow\)CH\(_3\)Br+ HBr is a combination reaction.

Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction.

CH\(_4\)+Br\(_2\) \(\rightarrow\)CH\(_3\)Br+ HBr

The above reaction is a combination reaction. In combination reaction, more than one reactant combine to form a product. Because new chemicals are created during combination reactions, they are often referred to as synthesis.

Therefore, the given reaction is a combination reaction.

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The concentration of H₂SO₄ solution is equal to 0.0480 M.

A neutralization reaction is described as a chemical reaction where acid and base react to produce respective salt and water. When a strong acid reacts with a strong base then the salt can be neutral.

When H₂SO₄ (a strong acid) reacts with NaOH, the resulting salt is Na₂SO₃  and water.

H₂SO₄  +  2 NaOH   →   Na₂SO₄  +   2H₂O

Given, the concentration of NaOH = 0.146 M

The volume of the NaOH = 14.85 ml = 0.01485 L

The number of moles of NaOH, n = M × V = 0.146 × 0.01485 = 0.00216 M

The volume of the  H₂SO₄ = 22.55 ml = 0.02255 L

The number of moles of  H₂SO₄, n = 0.00216/2 = 0.00108 mol

The concentration of H₂SO₄ =0.00108/0.02255 = 0.0480 M

Therefore, the molarity of  H₂SO₄ is 0.0480 M.

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There are 3 main states of matter: gas, solid, and liquid.

The solid is characterized by its rigid structure, its high density (it means that's there's a lot of mass per unit of volume). Also, it doesn't take the container's shape as a liquid does.

Then, the description corresponds to the solid state (option B).

Answer:

c

Explanation: