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9966 [12]
3 years ago
13

In the explosion of a hydrogen-filled balloon, 0.10 g of hydrogen reacted with 0.80 g of oxygen to form how many grams of water

vapor? (Water vapor is the only product.)
Express your answer using two significant figures.
Chemistry
1 answer:
lidiya [134]3 years ago
8 0

Mass of water vapor produced : 0.90 g

<h3>Further explanation</h3>

Given

0.10 g of hydrogen reacted with 0.80 g of oxygen

Required

mass of water vapor produced

Solution

Reaction

2H₂ + O ⇒ 2H₂O

If we refer to the law of conservation of mass which states that the mass before and after the reaction is the same, then the mass of water vapor formed as a product is:

mass reactants = mass products

mass of Hydrogen + mass of Oxygen = mass of water vapor

0.1 g + 0.8 g = 0.90 g

Or we can also solve<em> by using stochiometry</em> (using the concept of moles) to find the mass of water vapor

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Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30 10-8 at 700°C. 2 H2S(g) 2 H2(g) + S2(g) If
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Answer:

The equilibrium concentration of hydrogen gas is 0.0010 M.

Explanation:

The equilibrium constant of the reaction = K_c=9.30\times 10^{-8}

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[concentration]=\frac{moles}{volume (L)}

[H_2S]=\frac{0.31 mol}{4.1 L}=0.076 M

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Initially

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At equilibrium

(0.076-2x)                         2x     x

The expression of an equilibrium constant :

K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}

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A sucrose solution is prepared to a final concentration of 0.210 MM . Convert this value into terms of g/Lg/L, molality, and mas
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Answer:

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Step 2: Convert this value into terms of g/L

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Suppose we have a volume of 1.00L

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% MAss = (mass solute / mass solution)*100%

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