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kolezko [41]
2 years ago
5

Jason combines the two intermediate steps shown.

Chemistry
1 answer:
Gnom [1K]2 years ago
4 0
Hello ur answer will be D, hope this helps :)
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Many portable gas heaters and grills use propane, C3H8(g). Using enthalpies of formation, calculate the quantity of heat produce
Elodia [21]

Answer:  600 kJ

-

Explanation:

                        C₃H₈ (g)  + 5 O₂ (g) =============== 3 CO₂ (g)  +  4 H₂O (l)

Δ⁰Hf   kJ/mol       -104             0                                       -393.5         -285.8

Δ⁰Hcomb C₃H₈ = 3(-393.5) + 4 (-285.80) - (-104)  kJ/mol

Δ⁰Hcomb = 2219.70 kJ/mol

n= m /MW   MW c₃H₈ = 44.1 g/mol

n= 12 g/44.1 g/mol = 0.27 mol

then for 12 g the heat released will be

0.27 mol x 2219.70 kJ/mol =  600 KJ

4 0
3 years ago
Read 2 more answers
This green solution of chromium(III) can further be reduced by zinc metal to a blue solution of chromium(II) ions. Write the bal
Contact [7]

Answer:

Half-reactions:

Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻

Net ionic equation:

2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺

Explanation:

The Cr³⁺ is reduced to Cr²⁺:

<h3>Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>

Zn is oxidized to Zn²⁺:

<h3>Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>

Twice the reduction of Cr:

2Cr³⁺ + 2e⁻ → 2Cr²⁺

Now this reaction + Oxidation of Zn:

2Cr³⁺ + 2e⁻ + Zn → 2Cr²⁺ + Zn²⁺ + 2e⁻

<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>

6 0
2 years ago
How many atoms of S are there?
jonny [76]
Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1 . Because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.
4 0
3 years ago
Read 2 more answers
What is the percent composition of Br in CuBr3?
tekilochka [14]

Answer:

about 79% (79.04369332 to be exact)

Explanation:

Percent composition=(Molar mass of element x amount of it)/Molar mass of compound x 100

Br= 3 x 79.9/303.25 x100=79.04369332

6 0
3 years ago
Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t
iragen [17]

Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

    CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I                       0             0

C                     +S           +S

E                       S             S

The solubility product (Ksp) is:

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²

S = 7.9 × 10⁻⁵ M

<u>Solubility in 0.0120 M CoBr₂ (S')</u>

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

    CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I                       0           0.0240

C                     +S'           +S'

E                       S'            0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

8 0
3 years ago
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