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Tpy6a [65]
3 years ago
6

I’ll give brainliest please help thanks.

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

See Explanation

Explanation:

The equation of the reaction is;

C7H16(g) + 11O2(g) ---->7CO2(g) + 8H2O(g)

1a) Number of moles in 228 g of H2O = 228 g/18 g/mol = 12.67 moles

From the reaction equation;

11 moles of O2 yields 8 moles of H2O

x moles of O2 yields 12.67 moles of H2O

x = 11 * 12.67/8

x = 17.4 moles of O2

Since 1 mole of O2 occupies 22.4 L

17.4 moles of O2 occupies 17.4 moles * 22.4 L/1 mole = 389.76 L

1b) Molar mass of C7H16 = 100 g/mol

Number of moles in 300 g of C7H16 = 300 g/100g/mol = 3 moles

1 mole of C7H16 yields 7 moles of CO2

3 moles of C7H16 yields 3 * 7/1 = 21 moles of CO2

If 1 mole of CO2 occupies 22.4 L

21 moles of CO2 occupies 21 moles * 22.4 L/1 mole = 470.4 L of CO2

2) Number of moles in 202 L of H2 is obtained by;

1 mole of H2 occupies 22.4 L

x moles of  H2 occupies 202 L

x = 1 mole * 202 L/22.4 L =  9 moles of H2

From the reaction equation;

3 moles of H2 yields 2 moles of PH3

9 moles of H2 will yield 9 * 2/3 = 6 moles of PH3

Mass of 6 moles of PH3 = 6 moles of PH3 * 34 g/mol = 204 g of PH3

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The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
3 years ago
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