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guajiro [1.7K]
2 years ago
11

Part B

Chemistry
1 answer:
dezoksy [38]2 years ago
3 0

Answer:

5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=

Answer: 9.04 g of H2O

Explanation:

First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)

Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)

Use equation to get moles and plug given

5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O

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3 years ago
An airplane starts at rest and Excelerator down the runway for 20 seconds. At the end of the runway, its velocity is 80 m/s Nort
masha68 [24]
According to the formula you have given us to work with . . .

1). The airplane's acceleration is

(80 m/s north - zero) / (20 sec) = 4 m/sec^2 north

2).  For the cyclist:

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8 0
3 years ago
When the [H+] in a solution is 1.7 × 10−9 M, what is the pOH ? 8.80 5.23 14 8.60
kumpel [21]
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5 0
3 years ago
Read 2 more answers
NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.084 M in
kolbaska11 [484]
Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃]  = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
7 0
3 years ago
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