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zepelin [54]
3 years ago
15

4.13 moles of chromium (III) nitride weighs how much?

Chemistry
1 answer:
Leya [2.2K]3 years ago
3 0

Answer: 272.5915639999...

Explanation:

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Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o
Gnom [1K]

Answer: The empirical formula for the given compound is C_5H_7N

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

C_xH_yN_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of CO_2=21.363mg=21.363\times 10^3g=21363g

Mass of H_2O=6.125g=6.125\times 10^3g=6125g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, \frac{12}{44}\times 21363=5826.27g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, \frac{2}{18}\times 6125=680.55 of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = \frac{485.52}{97.73}=4.96\approx 5

For Hydrogen  = \frac{680.55}{97.73}=6.96\approx 7

For Nitrogen = \frac{97.73}{97.73}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is C_5H_7N_1=C_5H_7N

7 0
3 years ago
How many total elements are in marble?
Dahasolnce [82]

Answer:There are total 118 elements in world as about marbale

Explanation:

4 0
3 years ago
Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro
melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol

For the given chemical reaction:

3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)

  • <u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from \frac{3}{3}\times 0.252=0.252mol of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g

  • <u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from \frac{2}{3}\times 0.252=0.168mol of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

4 0
3 years ago
A sample of sulfur hexafluoride gas occupies a volume of 9.10 L at 198ÁC. Assuming that the pressure remains constant, what temp
alexandr1967 [171]
  As we know that
<span>V1/T1 = V2/T2 
V1 = 9.10 L 
T1 = 471 K 
V2 = 2.50 L 
T2 = 2.5 x 471 / 9.10 = 129.3 K 
T2 = 129.3 - 273 =
 -143.6 deg Celsiu
hope it helps</span>
5 0
3 years ago
Read 2 more answers
Which is an "enumerated" power of Congress?
fredd [130]

Answer:

In this section, we consider how several of the enumerated powers of Congress under the original Constitution have been interpreted. The Congress shall have Power To lay and collect Taxes, Duties, Imposts and Excises, to pay the Debts and provide for the common Defence and general Welfare of the United States.

6 0
2 years ago
Read 2 more answers
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