A student puts a piece of ice into a beaker of cold water. Which two Statements are true ?A. Thermal energy will move from air to the water. B. Thermal energy will move from the water to the ice. C.Thermal energy will move from the ice to the air . D. Thermal energy will move from the water to the air.

The stoichiometric table and the rate law for the given elementary liquid phase reaction have been constructed. The rate constant and activation energy have been calculated, and the type of reaction energy has been specified as endothermic.

Stoichiometric table in terms of concentrations:

The stoichiometric table for the given reaction can be constructed as follows:

A + 2B → products

                             A                                B                                products

Feed         0.3*Cf                        0.7*Cf                                       0

Exit              (0.3-0.3X)*C             (0.7-0.7X)*C                                0

Change         -0.3XC                      -0.7XC                                     0

Where:

Cf = Total feed concentration

C = Concentration inside reactor

X = Conversion of A

Rate of reaction of A:

The rate of the reaction can be expressed as:

rA = -1/2 * dCA/dt = k*C^2

where, CA is the concentration of A and k is the rate constant.

Since the reaction is elementary, the rate law is proportional to the concentrations of the reactants raised to their stoichiometric coefficients.

The rate of disappearance of A = rate of appearance of B

rB = -dCB/dt = 2*rA

Therefore, the rate of reaction of A can be expressed as:

rA = (0.7Cf - 0.7C)/V = k*C^2

Substituting values, we get:

rA = (0.710 - 0.70.7X)/5 = k(0.3 - 0.3*X)^2

Calculation of k and E:

The rate constant k can be calculated using the Arrhenius equation:

k = A * exp(-Ea/RT)

where A is the frequency factor, Ea is the activation energy, R is the gas constant and T is the temperature in Kelvin.

Assuming the activation energy is 50 kJ/mol, we can calculate the rate constant at the given temperature of 333 K:

k = 0.00717 * exp(-50000/(8.314*333)) = 0.0001504

The reaction energy can be determined by calculating the activation energy using the rate constant at two different temperatures. Assuming the rate constant at 323 K is 0.000098, we can solve for Ea:

ln(k2/k1) = Ea/R * (1/T1 - 1/T2)

ln(0.000098/0.0001504) = Ea/8.314 * (1/323 - 1/333)

Ea = 43775 J/mol

The positive value of the activation energy indicates that the reaction is endothermic.

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Answer:

\(m_{CO_2}=75.6gCO_2\)

Explanation:

Hello there!

In this case, according to the given information, it turns out mandatory for us to calculate the reacting moles of both C and O2 because we are given grams and pressure, temperature and volume, respectively:

\(n_C=36gC*\frac{1molC}{12gC}=3.0molC \\\\n_{O_2}=\frac{3.0atm*14L}{0.08206\frac{atm*L}{mol*K}*298K}=1.72molO_2\)

Thus, since C and O2 react in a 1:1 mole ratio, we infer C is in excess, and the grams of CO2 can be calculated with the moles of O2:

\(m_{CO_2}=1.72molO_2*\frac{1molCO_2}{1molO_2}*\frac{44.01gCO_2}{1molCO_2} \\\\ m_{CO_2}=75.6gCO_2\)

Best regards!

Answer:

\(mass \: number = n \: + p \\ 48 = 22 + p \\ p = 48 - 22 \\ = 26\)

It can be shown mathematical as

mass number = 48

neutrons = 22

Substitute the values

= 26 protons

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Answer:

THE ANSWER IS 1.167L

Explanation:

The metal X has an approximate molar mass of 42.36 g/mol and the metal is most likely calcium.

The molar mass of the metal carbonate XCO₃ and identify the metal X, we need to calculate the number of moles of XCO₃ and CO₂ using the given masses and molar masses.

The molar mass of CO₂ (carbon dioxide) is 12.011 g/mol (for carbon) + 2 * 15.999 g/mol (for oxygen) = 44.01 g/mol.

The number of moles of CO₂ can be calculated using the formula:

moles of CO₂ = mass of CO₂ / molar mass of CO₂

moles of CO₂ = 0.855 g / 44.01 g/mol

moles of CO₂ ≈ 0.01944 mol

Since the reaction stoichiometry is 1:1 between XCO₃ and CO₂, the number of moles of XCO₃ is also approximately 0.01944 mol.

molar mass of XCO₃ = mass of XCO₃ / moles of XCO₃

molar mass of XCO₃ = 2.012 g / 0.01944 mol

molar mass of XCO₃ ≈ 103.38 g/mol

The molar mass of XCO₃ is approximately 103.38 g/mol.

To determine the metal X:

molar mass of X = molar mass of XCO3 - molar mass of CO3

molar mass of X = 103.38 g/mol - (12.011 g/mol + 3 * 15.999 g/mol)

molar mass of X ≈ 42.36 g/mol

Metal X is most likely Calcium that has a molar mass of 40 g/mol

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Answer:

53.7 L

Explanation:

The volume occupied by 35.2 g of methane gas at 25 °C and 1.0 atm. is 55.9 L.

Given to us is temperature, pressure, universal gas constant, and mass of methane gas, we need to find the volume occupied by the gas.

To calculate the volume occupied by the methane gas, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = ideal gas constant (0.08206 Latm/(Kmol))

T = temperature (in Kelvin)

First, we need to convert the given temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15

T(K) = 25 °C + 273.15

T(K) = 298.15 K

Next, we calculate the number of moles of methane gas using its molar mass:

molar mass of CH₄ = 12.01 g/mol + 4(1.008 g/mol) = 16.04 g/mol

n = mass/molar mass

n = 35.2 g / 16.04 g/mol

n = 2.19 mol

Now we can substitute the values into the ideal gas law equation:

PV = nRT

V = (nRT) / P

V = (2.19 mol × 0.08206 Latm/(Kmol) ×298.15 K) / 1.0 atm

V = 55.9 L

Therefore, the volume occupied by 35.2 g of methane gas at 25 °C and 1.0 atm is 55.9 liters.

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Answer: a) current

Explanation:

After the reaction is complete, no nitrogen and no hydrogen molecules remain, and 8.00 x 1014 molecules of NH3 are produced.

In the equation, nitrogen and hydrogen react at a high temperature, in the presence of a catalyst, to produce ammonia, according to the balanced chemical equation:N2(g)+3H2(g)⟶2NH3(g)The coefficients of each molecule suggest that one molecule of nitrogen reacts with three molecules of hydrogen to create two molecules of ammonia.

So, to determine how many molecules of ammonia are produced when four nitrogen and nine hydrogen molecules are present, we must first determine which of the two reactants is the limiting reactant.

To find the limiting reactant, the number of moles of each reactant present in the equation must be determined.

Calculations:
Nitrogen (N2) molecules = 4Hence, the number of moles of N2 = 4/6.02 x 1023 mol-1 = 6.64 x 10-24 mol
Hydrogen (H2) molecules = 9Hence, the number of moles of H2 = 9/6.02 x 1023 mol-1 = 1.50 x 10-23 mol

Now we have to calculate the number of moles of NH3 produced when the number of moles of nitrogen and hydrogen are known, i.e., mole ratio of N2 and H2 is 1:3.

The mole ratio of N2 to NH3 is 1:2; thus, for every 1 mole of N2 consumed, 2 moles of NH3 are produced.
The mole ratio of H2 to NH3 is 3:2; thus, for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
From these mole ratios, it can be observed that the limiting reactant is nitrogen.

Calculation for NH3 production:
Nitrogen (N2) moles = 6.64 x 10-24 moles
The mole ratio of N2 to NH3 is 1:2; therefore, moles of NH3 produced is 2 × 6.64 × 10−24 = 1.33 × 10−23 moles.

Now, to determine how many molecules of NH3 are produced, we need to convert moles to molecules.
1 mole = 6.02 x 1023 molecules
Thus, 1.33 x 10-23 moles of NH3 = 8.00 x 1014 molecules of NH3 produced.

To find the amount of each reactant remaining after the reaction is complete, we must first determine how many moles of nitrogen are consumed, then how many moles of hydrogen are consumed, and then subtract these from the initial number of moles of each reactant.

The moles of nitrogen consumed = 4 moles × 1 mole/1 mole N2 × 2 mole NH3/1 mole N2 = 8 moles NH3
The moles of hydrogen consumed = 9 moles × 2 mole NH3/3 mole H2 × 2 mole NH3/1 mole N2 = 4 moles NH3
Thus, the moles of nitrogen remaining = 6.64 × 10−24 mol – 8 × 2/3 × 6.02 × 10^23 mol-1 = 5.06 × 10−24 mol
The moles of hydrogen remaining = 1.50 × 10−23 mol – 4 × 2/3 × 6.02 × 10^23 mol-1 = 8.77 × 10−24 mol

Finally, the number of molecules of each reactant remaining can be calculated as follows:
Number of N2 molecules remaining = 5.06 × 10−24 mol × 6.02 × 10^23 molecules/mol = 3.05 × 10−1 molecules ≈ 0 molecules
Number of H2 molecules remaining = 8.77 × 10−24 mol × 6.02 × 10^23 molecules/mol = 5.28 × 10−1 molecules ≈ 0 molecules.

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Explanation

Halogens => Cl, Br, I, F => exist as diatomic molecules at room temperature and atmospheric conditions.

F2 and Cl2 are gases

Br2 is liquid

I2 is solid

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Answer:

\(m_{O_2}=120gO_2\)

Explanation:

Hello!

In this case, since the decomposition of potassium chlorate is:

\(2KClO_3\rightarrow 2KCl+3O_2\)

We can see a 2:3 mole ratio between potassium chlorate and oxygen (molar mass 32.0 g/mol), thus, via stoichiometry, we compute the mass of oxygen that are produced by the decomposition of 2.50 moles of this reactant:

\(m_{O_2}=2.50molKClO_3*\frac{3molO_2}{2molKClO_3} *\frac{32.0gO_2}{1molO_2}\\\\m_{O_2}=120gO_2\)

Best regards!

Answer:

Final pressure is 1.42atm

Explanation:

Based on Gay-Lussac's law, pressure of a gas is directely proportional to its absolute temperature. The equation of this law is:

P₁T₂ = P₂T₁

Where P is pressure and T is absolute temperature of 1, initial state and 2, final state of the gas.

In the problem, initial conditions are Standard Temperature and Pressure, STP, that are 1 atm and 273.15K.

If the final temperature is 115°C = 388.15K (115°C + 273.15 = 388.15K), using Gay-Lussac's law:

P₁T₂ = P₂T₁

1atmₓ388.15K = P₂ₓ273.15K

1.42atm = P₂

Answer: One methane molecule, CH4, contains one carbon atom and four hydrogen atoms.

Explanation:

Answer:

1. 0.28M

2. 24mL

3. 0.002295moles of NaOH.

Explanation:

1. The 6.0M HCl solution is diluted from 35mL to 750mL, that is:

750mL / 35mL = 21.43 times

The 6.0M HCl solution is diluted 21.43 times, the final concentration is:

6.0M / 21.43 = 0.28M

2. First, we need to convert grams of CuNO3 to moles and then, as molarity is moles per liter we can solve for volume of solution in liters and convert it to mililiters:

Moles CuNO3 -Molar mass 124.9g/mol:

2.49g * (1mol / 124.9g) = 0.020 moles of CuNO3

Volume is:

0.020 moles * (1L / 0.830mol) = 0.0240L * (1000mL / 1L) = 24mL

3. In 13.5mL = 0.0135L of 0.17M NaOH the moles are:

0.0135L * (0.17mol / L) = 0.002295moles of NaOH

Answer:

9.10% dextrose

Explanation:

To find the mass percent, you need to use the following equation:

                               mass (g) of solute
Mass Percent  =  ---------------------------------  x  100%
                              mass (g) of solution

You can plug the given values into the equation and solve to find the mass percent of dextrose. But first, you need to calculate the mass of the solution.

Mass (solute): 25.0 g

Mass (solution): 250.0 g + 25.0 g = 275 g

                                25.0 grams
Mass Percent  =  ------------------------  x  100%
                                 275 grams

Mass Percent  =  0.0909 x 100%

Mass Percent  =  9.10%

A solution of dextrose contains 25.0 g solute in 250.0 g water. 9.10% is the percentage by mass of dextrose in this solution.

A component's concentration in a combination or compound can be expressed as a mass percent, also known as a weight percent. It shows how much of the total mass of the solution or mixture is made up of the solute's (component's) mass. In chemistry, mass percent is frequently employed and stated as a percentage.

Mass Percent  =     mass (g) of solute/       mass (g) of solution x  100%

Mass (solute): 25.0 g

Mass (solution): 250.0 g + 25.0 g = 275 g

Mass Percent  =25.0 grams /    275 grams x  100%

Mass Percent  =  0.0909 x 100%

Mass Percent  =  9.10%

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Answer:

1.014×10²³ atoms

Explanation:

Given data:

Volume of tungsten = 1.6 cm³

Density of tungsten = 19.35 g/cm³

Number of atoms of tungsten = ?

Solution:

First of all we will calculate the mass of tungsten.

density = mass/volume

by putting values,

19.35 g/cm³ = mass/1.6 cm³

Mass = 19.35 g/cm³× 1.6 cm³

Mass = 30.96 g

Number of moles of tungsten:

Number of moles = mass/molar mass

Number of moles = 30.96 g/ 183.84 g/mol

Number of moles = 0.1684 mol

1  mole contain 6.022×10²³ atoms.

0.1684 mol × 6.022×10²³ atoms / 1mol

1.014×10²³ atoms

Answer:

If by gas you mean oxygen it is O.

Answer:

Explanation:

liquid

Answer:

A. 1 J=1kg•m^2/s^2

Explanation:

Energy refers to the capacity to do work. According to the International System of units (SI units), energy is measured in Joules.

Energy is represented by the force applied over a distance. Force is measured in Newton (N) and distance in metres (m). Hence, energy is Newton × metre (N.m)

Newton is derived from the SI units of mass (Kilograms), and acceleration (metres per seconds^2) i.e Kg.m/s^2, since Force = mass × acceleration.

Since; Energy = Newton × metres

If Newton = Kg.m/s^2 and metres = m

Energy (J) will therefore be; Kg.m/s^2 × m

1J = Kg.m^2/s^2

Explanation:

Bohr's model of the hydrogen atom is based on three postulates: (1) an electron moves around the nucleus in a circular orbit, (2) an electron's angular momentum in the orbit is quantized, and (3) the change in an electron's energy 

Answer:

please give me brainlist and follow

Explanation:

Substance in water + 3 mL Benedict's solution, then boil for few minutes and allow to cool. The common disaccharides lactose and maltose are directly detected by Benedict's reagent because each contains a glucose with a free reducing aldehyde moiety after isomerization.

Glucose and maltose give a positive Benedict test because they have a free carbonyl group to undergo oxidation.

Explanation:

So, from this, we can conclude that glucose and maltose give a positive Benedict test because they have a free carbonyl group to undergo oxidation.

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CuI2 is not a known compound. Copper compounds typically have different oxidation states for copper, resulting in various compound names.

Copper(II) oxide (CuO): It is a black solid compound where copper is in the +2 oxidation state. It is commonly used as a pigment and in catalytic reactions.

Copper(II) sulfate (CuSO4): It is a blue crystalline compound in which copper is in the +2 oxidation state. It is used in various applications such as agriculture, electroplating, and as a laboratory reagent.

Copper(I) oxide (Cu2O): It is a red crystalline compound in which copper is in the +1 oxidation state. It is used as a pigment, in solar cells, and as a catalyst.

Copper(II) chloride (CuCl2): It is a greenish-brown solid compound in which copper is in the +2 oxidation state. It is utilized in various chemical processes, including etching and catalyst synthesis.

Copper(II) nitrate (Cu(NO3)2): It is a blue crystalline compound where copper is in the +2 oxidation state. It is commonly used in the production of catalysts, as a coloring agent, and in electroplating.

These are just a few examples of copper compounds with different oxidation states and properties. It's important to note that the compound CuI2 mentioned in the question, if it exists, would be an exception to the typical nomenclature for copper compounds.

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Answer:

it is 7 and xs hope this helps :) !

Explanation:

Answer:

Explanation:

The water in a garden hose flows at a rate of 10.0 liters per minute. Therefore, 158 gal/hr is the flow rate in gallons per hour.

Flow rate is the amount of liquid that travels in a given length of time. Furthermore, the flow rate is affected by the channel through which the liquid is travelling, the area of the pipe, as well as the velocity of something like the liquid. Furthermore, the formula is. Fluid dynamic rate = pipe or channel area x liquid velocity.

The volume flow value is the rate of liquid (water flow rate formula) passing through every place in an area over time.

1 gal = 3.79 L

60 min = 1 hr

10.0 L/min x 60 min/hr x 1 gal/3.79 L = 158 gal/hr

Therefore, 158 gal/hr is the flow rate in gallons per hour.

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Answer:

66 grams of ice would have to melt to lower the temperature of 352 mL of water from 15 °C to 0 °C.

Explanation:

To calculate the amount of ice that would have to melt to lower the temperature of 352 mL of water from 15 °C to 0 °C, we need to use the formula:

Q = m_water * c_water * ΔT_water + m_ice * Lf

where,
Q = the amount of heat transferred,
m_water = the mass of water, c_water is the specific heat capacity of water,
ΔT_water = the change in temperature of water, m_ice = the mass of ice,
Lf = the specific latent heat of fusion of ice.

First, let's calculate the amount of heat transferred to the water:

Q = m_water * c_water * ΔT_water
Q = 352 g * 1.0 cal/(g*°C) * (15-0) °C
Q = 5,280 cal

Next, we can use the specific latent heat of fusion of ice, which is 80 cal/g, to calculate the amount of heat required to melt the ice:

Q = m_ice * Lf
Q = m_ice * 80 cal/g
m_ice = Q / Lf
m_ice = 5,280 cal / 80 cal/g
m_ice = 66 g

Answer:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases.

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Explanation:

Hello,

In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.

With the aforementioned, we can conclude that the chemical reaction:

\(N_2O_4(g) \rightleftharpoons 2 NO_2(g)\)

Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:

\(G=H-TS\)

As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:

\(K=exp(-\frac{\Delta G}{RT} )\)

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Regards.