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3 years ago

Find the mass of 3.50x10^23 molecules of nitrogen gas

Chemistry

1 answer:

Zarrin [17]3 years ago

4 0

Answer:

16.27 g.

Explanation:

It is known that every 1.0 mol of a compound contains Avogadro's no. of molecules (6.022 x 10²³).

Using cross multiplication:

1.0 mol of N₂ gas contains → 6.022 x 10²³ molecules.

??? mol of N₂ gas contains → 3.50 x 10²³ molecules.

∴ The no. of moles of N₂ gas = (1.0 mol)(3.50 x 10²³ molecules)/(6.022 x 10²³ molecules) = 0.5812 mol.

∴ The mass of 3.50 x 10²³ molecules of N₂ gas = (no. of moles)*(molar mass) = (0.5812 mol)(28.0 g/mol) = 16.27 g.

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Caffeine, a stimulant in coffee and tea, has a molar mass of 194.19 g/mol and a mass percentage composition of 49.48% C, 5.19% H

lozanna [386]

Answer : The molecular formula of a caffeine is, $C_8H_{10}N_4O_2$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.48g}{12g/mole}=4.12moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.19g}{1g/mole}=5.19moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{4.12}{1.03}=4$

For H = $\frac{5.19}{1.03}=5.03\approx 5$

For N = $\frac{2.06}{1.03}=2$

For O = $\frac{1.03}{1.03}=1$

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_4H_5N_2O_1=C_4H_5N_2O$

The empirical formula weight = 4(12) + 5(1) + 2(14) + 16 = 97 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{194.19}{97}=2$

Molecular formula = $(C_4H_5N_2O)_n=(C_4H_5N_2O)_2=C_8H_{10}N_4O_2$

Therefore, the molecular of the caffeine is, $C_8H_{10}N_4O_2$

5 0

4 years ago

Use IUPAC rules to name this compound. A) nitrogen iodide B) nitrogen triiodide C) mononitrogen iodide D) mononitrogen triiodide

Mariulka [41]

Answer:

Name of the compound is Nitrogen triiodide.

Explanation:

According to the IUPAC rules, to naming of the compound the following formula can be applied.

Prefix + Name of first element + Base name element of second element + Suffix.

The given compound - $NI_{3}$

Name of first element- Nitrogen

Base name element of second element - Iodine

Suffix = 3 = tri

Here, iodine is in ionic form therefore, it becomes iodide. and then suffix will be added in front of the halogen.

Therefore, name of the compound will be Nitrogen triiodide..

3 0

3 years ago

Read 2 more answers

Hello!

blondinia [14]

Answer:

$[CO]=[Cl_2]=0.01436M$

$[COCl_2]=0.00064M$

Explanation:

Hello there!

In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:

$K=\frac{[CO][Cl_2]}{[COCl_2]}$

Which can be written in terms of x, according to the ICE table:

$0.32=\frac{x^2}{0.015M-x}$

Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:

$[CO]=[Cl_2]=0.01436M$

$[COCl_2]=0.015M-0.01436M=0.00064M$

Regards!

5 0

3 years ago

Which of the following equations is balanced?

Verizon [17]

answer is B

because both sides have equal numbers of atomos

7 0

3 years ago

If the mass of monomer 105 atomic mass units, find the mass of a polymer contianing 122 monomers

sukhopar [10]

12810 atomic mass units
Since a monomer is the basic building block of a polymer, we just need to multiply the mass of the monomers by the number of monomers used. So 105 atomic mass units * 122 = 12810 atomic mass units

3 0

3 years ago