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Shalnov [3]
3 years ago
13

To calculate the enthalpy change for the reaction, 2CO (g) + O2 (g) Imported Asset 2 CO2 (g), you can use ΔHf0 values for each r

eactant and product. Which setup shows how the enthalpy change should be calculated?
ΔH0reaction = [ΔHf0 CO2(g)] + [ΔHf0 CO(g) - ΔHf0 O2(g)]

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

ΔH0reaction = [ΔHf0 CO(g) + ΔHf0 O2(g)] - [ΔHf0 CO2(g)]

ΔH0reaction = [ΔHf0 CO(g) - ΔHf0 O2(g)] + [ΔHf0 CO2(g)]
Chemistry
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

Explanation:

Chemical equation:

CO + O₂   →  CO₂

Balanced chemical equation:

2CO + O₂   →  2CO₂

The standard enthalpy for the formation of CO = -110.5 kj/mol

The standard enthalpy for the formation of O₂  = 0  kj/mol

The standard enthalpy for the formation of CO₂  = -393.5 kj/mol

Now we will put the values in equation:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]

ΔH0reaction = -283 kj/mol

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scoundrel [369]
I think the correct answer would be NH4ClO4. It would produce a basic solution since as it dissociates into ions OH ions are being formed as the concentration of this increases, the pH of the solution would increase as well making it basic. Hope this helps.
5 0
3 years ago
he equilibrium constant, Kc , for the following reaction is 7.00×10-5 at 673 K.NH4I(s) NH3(g) + HI(g)If an equilibrium mixture o
frosja888 [35]

<u>Answer:</u> The number of moles of HI in the solution is 1.24\times 10{-3} moles.

<u>Explanation:</u>

We are given:

K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L

To calculate the concentration of a substance, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     ......(1)

  • Concentration of ammonia:

[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L

  • Concentration of ammonium iodide:

[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L

For the given chemical reaction:

NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)

The expression of K_c for above equation follows:

K_c=\frac{[HI][NH_3]}{[NH_4I]}

Putting values in above equation, we get:

7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}

[HI]=2.53\times 10^{-4}

Calculating the moles of hydrogen iodide by using equation 1, we get:

2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}

Hence, the number of moles of HI in the solution is 1.24\times 10{-3} moles.

3 0
3 years ago
A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 667 torr. Use Boyle's law to calculate the pres
zaharov [31]

Answer:

The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)

Explanation:

Step 1: Data given

The temperature of a gas = 25.0°C

AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.

The volume reduces to 7.88 L but the temperature stays constant.

Step 2: Boyle's law

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

 ⇒ Since the temperature stays constant, we can simplify to:

P1*V1 = P2*V2

⇒ with P1 = the initial pressure 667 torr

⇒ with V1 = the initial volume = 10.0 L

⇒ with P2 = the final pressure = TO BE DETERMINED

⇒ with V2 = the final volume = 7.88L

P2 = (P1*V1)/V2

P2 = (667*10.0)/7.88

P2 = 846 torr

The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)

6 0
3 years ago
How many moles of sucrose<br> are in 5.25x1029 sucrose<br> molecules? (in scientific notation)
Lorico [155]

Answer:

8.72 × 10^5 moles

Explanation:

To find the number of moles in 5.25 x 10^29 molecules of sucrose, we divide the number of molecules by Avagadro constant (6.02 × 10²³ molecules). That is;

no. of moles = no. of molecules ÷ 6.02 × 10²³ molecules

In this case of sucrose, no of moles contained is as follows;

5.25 × 10^29 ÷ 6.02 × 10²³

5.25/6.02 × 10^ (29-23)

0.872 × 10^6

= 8.72 × 10^5 moles

5 0
3 years ago
Tuliskan persamaan tetapan kesetimbangan untuk reaksi-reaksi berikut a. Fe3+(aq) + SCN– (aq) ↔ FeSCN3+(aq) b. 3Fe(s) + 4H2O(g) ↔
nevsk [136]

Answer:

a. K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}

b. K_p = \dfrac{[H_2]^4}{[H_2O]^4}

Explanation:

Untuk semua jenis reaksi umum:

aA + bB\iff cC + dD

Konstanta kesetimbangan K_c = \dfrac{[C]^c [D]^d}{[A]^a[B]^b}

Dari pertanyaan yang diberikan:

a. Fe3^+_{(aq)} + SCN^-_{ (aq)} \iff FeSCN^{3+}_{(aq) }

Konstanta kesetimbangan:

K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}

b. 3Fe_{(s)} + 4H_2O_{(g)} \iff Fe_3O_4_{(s)} + 4H_{2(g)}

Konstanta kesetimbangan untuk tekanan parsial K_p

K_p = \dfrac{[H_2]^4}{[H_2O]^4}

Karena Fe3O4 (s) hadir sebagai padatan.

8 0
3 years ago
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