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natima [27]
2 years ago
15

The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C u

pper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed. is halved. has its sign changed. is unchanged.
Chemistry
1 answer:
GenaCL600 [577]2 years ago
7 0

Answer:

the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

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If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________. If the initial concen
Nikolay [14]

This is an incomplete question, here is a complete question.

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.

The reaction

2NOBr(g)\rightarrow 2NO(g)+Br_2(g)

It is a second-order reaction with a rate constant of 0.80 M⁻¹s⁻¹ at 11 °C.

Answer : The concentration of after 9.0 seconds is, 0.00734 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80 M⁻¹s⁻¹

t = time taken  = 142 second

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.0440 M

Putting values in above equation, we get:

0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)

[A]=0.00734M

Hence, the concentration of after 9.0 seconds is, 0.00734 M

5 0
2 years ago
C5H12, pentane, is a liquid at room temperature.
Art [367]

Answer:

C)We cannot be sure unless we find out its boiling point.

Explanation:

It is necessary to clearly explain here that simply observing two compounds of the same homologous series irrespective of how close they may be in the series will not give us the faintest idea regarding which one will be a liquid, solid or gas at room temperature.

However, to determine whether an unknown substance will be a liquid at room temperature, then its important to measure its boiling point. If the boiling point is above room temperature, and the melting point is below room temperature, the compound is a liquid. If the boiling point of the unknown substance is below room temperature, it is a gas.

It is now safe to conclude that cannot decide on the state of matter in which a compound exists unless we know something about its boiling point, not merely looking closely at the properties of its neighbouring compounds in the same homologous series

8 0
2 years ago
one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40
Ronch [10]

<u>Answer:</u> The volume of chlorine gas produced in the reaction is 2.06 L.

<u>Explanation:</u>

  • <u>For potassium permanganate:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol

  • <u>For hydrochloric acid:</u>

To calculate the moles of hydrochloric acid, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol

  • For the reaction of potassium permanganate and hydrochloric acid, the equation follows:

2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = \frac{2}{16}\times 0.27=0.033moles of potassium permanganate.

As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = \frac{5}{16}\times 0.27=0.0843moles of chlorine gas.

  • To calculate the volume of gas, we use the equation given by ideal gas equation:

PV=nRT

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = 0.0820\text{ L atm }mol^{-1}K^{-1}

T = temperature of the gas = 40^oC=[40+273]K=313K

Putting values in above equation, we get:

1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

3 0
2 years ago
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WARRIOR [948]

Answer:

20.2 g silver placed in 21.6 mL of water and 20.2 g copper placed in 21.6 mL of water

Explanation:

Since the density formula is mass divided volume, you can calculate the density of each of the components of the group and see which of all the groups are equal

In the group

density Ag=20.2 g silver/21.6ml H2O=0.9352g/ml

density Cu=20.2 g copper/21.6ml H2O=0.9352 g/ml

6 0
2 years ago
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