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alekssr [168]
3 years ago
12

How many milliliters of manganese metal, with a density of 7.43 g/mL, would be needed to produce 21.7 grams of hydrogen gas in t

he single-replacement reaction below? Show all steps of your calculation as well as the final answer. Mn + H2O → MnO + H2 (5 points)
Chemistry
1 answer:
Alla [95]3 years ago
3 0
First compute the number of grams of manganese metal required to make 21.7 grams of H2. 
21.7 g H2 x (1 mole H2/ 2 g H2) x (1 mole Mn/1 mol H2) x (55 grams Mn/1 mol Mn) = 596.75 grams 
Now density = mass/volume 
7.43 = 596.75/volume 
volume = 596.75/7.43 = 80.31 mL 
80.31 mL is the amount of manganese needed.
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A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de
jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

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Molar mass of CH3COOH,

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= 2*12.01 + 4*1.008 + 2*16.0

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use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

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density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

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4 years ago
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4 0
3 years ago
Read 2 more answers
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2 years ago
What is the percent composition of sodium (na) in baking soda
vagabundo [1.1K]

Element                                          Symbol                                                   Mass Percent

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