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3 years ago

(6.0×10^4)(3.1×10^-1)

Chemistry

1 answer:

yKpoI14uk [10]3 years ago

6 0

(6.0×10^4) = 60,000
(3.1×10^-1) = 0.31

60,000 × 0.31 = 18,600

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Answer:

$K_{goal}=1.793*10^{-33}$

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If we add above reaction we will get:

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Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

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how many moles of a nonvolatile, nonelectrolyte solute are required to lower the freezing point of 1000 grams of water by 5.58

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Answer:

Explanation:

Using freezing point depression formula,

ΔTemp.f = Kf * b * i

Where,

ΔTemp.f = temp.f(pure solvent) - temp.f(solution)

b = molality

i = van't Hoff factor

Kf = cryoscopic constant

= 1.86°C/m for water

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= 3.00 mol/kg

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= (3.00 x 1)

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Explanation:

Data given:

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mass of water = 26.6 grams

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from the principle of:

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