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3 years ago

How many grams of n2f4 can be produced from 225g f2?

1 answer:

8 0

The answer is 615.91 grams of <span>n2f4

Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
=615.91 grams

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2 years ago

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What is the best description of the energy stored in a stretched or compressed object.

oksian1 [2.3K]

Answer:

Elastic Potential Energy

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3 years ago

(A tablet containing calcium carbonate and fillers with a mass of 1.631 g was dissolved in HCl. After the fillers were filtered

FrozenT [24]

Answer:

1) 0.825 grams is the mass of pure calcium carbonate product collected at the end of the experiment.

2) The mass % calcium carbonate in the tablet is 50.58%.

3) The mass % of calcium in calcium carbonate is 40.00 %.

4) Mass of calcium in a tablet is 0.33 g.

Explanation:

Mass of tablet = 1.631 g

Mass of the watch glass = 46.719 g

Mass of the watch glass + mass of calcium carbonate = 47.544 g.

Mass of calcium carbonate = 47.544 g - 46.719 g = 0.825 g

0.825 grams is the mass of pure calcium carbonate product collected at the end of the experiment.

Mass of calcium carbonate = 0.825 g

Percentage of calcium in tablet:

$\%=\frac{\text{Mass of calcium carbonate }}{\text{Mass of tablet}}\times 100$

$\%=\frac{0.825 g}{1.631 g}\times 100=50.58\%$

The mass % calcium carbonate in the tablet is 50.58%.

Percentage of calcium in calcium carbonate :

Molar mass of calcium carbonate = 100 g/mol

Atomic mass of calcium = 40 g/mol

Percentage of calcium in calcium carbonate :

$\%=\frac{\text{atomic mass of calcium}}{\text{Molar mass of}CaCO_3}\times 100$

$\%=\frac{40 g/mol}{100 g/mol}\times 100=40.00\%$

The mass % of calcium in calcium carbonate is 40.00 %.

Mass of calcium carbonate = 0.825 g

The mass % of calcium in calcium carbonate is 40.00 %.

Mass of calcium in a tablet : x

$40.00\%=\frac{\text{Mass of calcium in a tablet}}{0.825 g}\times 100$

$x=40.00\times \frac{0.825 g}{100}=0.33 g$

Mass of calcium in a tablet is 0.33 g.

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