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astra-53 [7]
3 years ago
7

In one compound of nitrogen and oxygen, 0.615 grams of nitrogen combines with 0.703 grams of oxygen. In another 1.27 grams of ni

trogen combines with 2.90 grams of oxygen. Show how these data illustrate the law of multiple proportions.
Chemistry
1 answer:
Orlov [11]3 years ago
4 0

Answer:

Answer in explanation

Explanation:

In the first case, we divide each of the masses by the respective atomic masses:

N =0.615/14 = 0.043928571428571

O = 0.703/16 = 0.0439375

It can be seen here that the values are similar, hence the formula is NO

now let us look at the second data set:

N = 1.27/14 = 0.090714285714286

O = 2.9/16 = 0.18125

We now divide by the smallest

N = 090714285714286/090714285714286 = 1

O = 0.18125/090714285714286 = 2

The formula here is thus NO2.

It can be seen that there are different oxides of nitrogen here which clearly indicates the law of multiple proportion.

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Ni(OH)2+H2SO4=NiSO4+2H2O
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5 0
3 years ago
For the reaction 2H2 + O2 --> 2H2O, how many grams of oxygen are needed to react 3 moles of hydrogen?
mr Goodwill [35]

Answer:

48 grams

Explanation:

The chemical equation for the reaction is the following:

2 H₂ + O₂ → 2 H₂O

That means that 2 moles of H₂ react with 1 mol of O₂ to produce 2 moles of H₂O. We convert the moles of oxygen (O₂) by using the molecular weight (MW) as follows:

MW(O₂) = 16 g/mol x 2 = 32 g/mol

mass of O₂ = 1 mol x 32 g/mol = 32 g

So, we have the following stoichiometric ratio: 32 g O₂/2 moles H₂. We have 3 moles of hydrogen (H₂), so we multiply the moles by the stoichiometric ratio to calculate how many grams are needed:

3 moles H₂ x 32 g O₂/2 moles H₂ = 48 g O₂

<em>Therefore, 48 grams of O₂ are needed to react with 3 moles of H₂.</em>

4 0
3 years ago
The volume of a gas held at constant temperature varies indirectly as the pressure of the gas. If the volume of a gas is 1200 cu
kvasek [131]

Answer:

Volume of the gas at 500 mm Hg pressure is 960 cm^{3}

Explanation:

Let's assume the gas behaves ideally.

According to combined gas law for an ideal gas-

                  \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

Where P_{1} and  P_{2} are initial and final pressure of the gas respectively.  V_{1} and  V_{2} are initial and final volume of the gas respectively.  T_{1} and  T_{2} are initial and final temperature of the gas in kelvin respectively.

Here T_{1} = T_{2}, V_{1} = 1200 cm^{3}, P_{1} = 400 mm Hg, P_{2} = 500 mm Hg

So, V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}

Hence V_{2}=\frac{400\times 1200}{500}[tex]cm^{3}=960 cm^{3}[/tex]

Hence volume of the gas at 500 mm Hg pressure is 960 cm^{3}

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4 years ago
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How many elements are in 2CaCO3 and C8H10N402
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In 2CaCO3 there is 3 elements.

In C8H10N402 there is 3 elements.

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