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Kipish [7]
3 years ago
15

The combustion of propane (C3H8) produces CO2 and H2O: C3H8 (g) 5O2 (g) → 3CO2 (g) 4H2O (g) The reaction of 2.5 mol of O2 will p

roduce ________ mol of H2O.
Chemistry
1 answer:
olganol [36]3 years ago
3 0

Answer: 2 moles of water.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The combustion of propane is represented as :

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

5 moles of O_2  produces = 4 moles of H_2O

2.5 moles of O_2  will produce=\frac{4}{5}\times 2.5=2 moles of H_2O.

Thus 2 moles of water are produced by 2.5 moles of oxygen.

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Under what circumstances may a health insurer charge a higher premium to a woman with a genetic disposition to breast cancer? a)
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Answer:

D

Explanation:

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8 0
2 years ago
Consider a solution that is 1.3×10−2 M in Ba2+ and 2.0×10−2 M in Ca2+. Ksp(BaSO4)=1.07×10−10 Ksp(CaSO4)=7.10×10−5 If sodium sulf
skad [1K]

Answer:

Explanation:

Ksp(BaSO4)=1.07×10−10

BaSO₄ → Ba²⁺ + SO₄²⁻

1.07×10⁻¹⁰ = ( Ba²⁺) × ( SO₄²⁻)

but Ba²⁺ = 1.3×10⁻² M

1.07×10⁻¹⁰  = 1.3×10⁻² M × ( SO₄²⁻)

( SO₄²⁻)  = 1.07×10⁻¹⁰  / 1.3×10⁻² = 0.823 × 10⁻⁸ M

while Ksp(CaSO4)=7.10×10−5

CaSO₄ → Ca²⁺ + SO₄²⁻

7.10×10⁻⁵ = 2.0×10⁻² × ( SO₄²⁻)

( SO₄²⁻)  = 7.10×10⁻⁵  /  2.0×10⁻² = 3.55 × 10⁻³ M

comparing the concentration of sulfate ions, Ba²⁺ cation will precipitate first because the Ba²⁺ requires 0.823 × 10⁻⁸ M sodium sulfate which less compared the about needed by CaSO₄

7 0
3 years ago
TRUE OR FALSE 3D printing waste less material than traditional manufacturing.
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3 years ago
Phosgene, a poisonous gas, when heated will decompose into carbon monoxide and chlorine in a reversible reaction: COCl2 (g) <
Advocard [28]

Answer:

8.08 × 10⁻⁴

Explanation:

Let's consider the following reaction.

COCl₂(g) ⇄ CO (g) + Cl₂(g)

The initial concentration of phosgene is:

M = 2.00 mol / 1.00 L = 2.00 M

We can find the final concentrations using an ICE chart.

     COCl₂(g) ⇄ CO (g) + Cl₂(g)

I       2.00            0            0

C        -x             +x           +x

E    2.00 -x          x             x

The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.

The concentrations at equilibrium are:

[COCl₂] = 2.00 -x = 1.96 M

[CO] = [Cl₂] = 0.0398 M

The equilibrium constant (Keq) is:

Keq = [CO].[Cl₂]/[COCl₂]

Keq = (0.0398)²/1.96

Keq = 8.08 × 10⁻⁴

4 0
3 years ago
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