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2 years ago

What mass of hydrogen contains the same number of atoms as 7.00 g of nitrogen

1 answer:

7 0

Molecular mass of nitrogen, N2 = 2*14 = 28
molecular mass of hydrogen, H2=2*1 = 2

Molecules of both elements contain 2 atoms, so the ratio of molecules is the same as the ratio of atoms.

From the molecular masses above,
ratio of number of molecules = ratio of molecular masses, therefore
7g N2 : x g H2 = 28:2
cross multiply:
x=7*2/28=0.5 g of Hydrogen has the same number of atoms as 7 g of nitrogen (at room temperatures)

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Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample

goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=2:2=1:1$

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=3:3=1:1$

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.

4 0

2 years ago

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

Diano4ka-milaya [45]

The balanced chemical reaction is given as follows:

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:

2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2

0.361 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2

83.6 kg KClO3 (1000g / 1kg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2

22.5 mg KClO3 (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2

4 0

2 years ago

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The morning temperature in a city is 41°F. If a sunny, mild day is forecast, which temperature is most likely for 2:00 p.m.?

romanna [79]

Answer:

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2 years ago

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AleksAgata [21]

2-A
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2 years ago

Which of these is a chemical property? color, compressibility, heat of combustion, malleability

Flauer [41]

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2 years ago

Read 2 more answers