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mina [271]
2 years ago
6

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

Chemistry
1 answer:
tekilochka [14]2 years ago
7 0

Answer: There are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

So, the number of atoms present in 0.67 moles are as follows.

0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms

Thus, we can conclude that there are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

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1) If you have 2.6 moles of iron (III) oxide, how many molecules of iron (III)
Ugo [173]

Answer:

234

Explanation:

so 3 x 3 x 26 =234

8 0
3 years ago
I need help please ASAP
Yuliya22 [10]

Answer:

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6 0
2 years ago
Calculate the molecular (formula) mass of each compound: (a) iron(ll) acetate tetrahydrate; (b) sulfur tetrachloride; (c) potass
Nutka1998 [239]

Answer:

a) Iron(ll) acetate tetrahydrate: 245,68 g/mol

b) Sulfur tetrachloride: 173,87 g/mol

c) Potassium ermanganate 158,034 g/mol

Explanation:

To solve this kind of exercises you must look for the number of atoms in each molecule first, then look on the periodic table the atom weight and the multiply the atom weight times the quantity of each atom. For instance:

The molecule of Iron(II) acetate itetrahydrate is (CH3COO)2Fe•4H2O, it means that you have:

2 atoms of carbon times the atomic weight of C (12.00g/mol)= 24g

14 atoms of Hidrogen times the atomic weight of H (1,00g/mol)= 14g

6 atoms of Oxigen times the atomic weight of O (16,0g/mol)= 96

2 atoms of Iron times the atomic weight of Fe (55,84g/mol)= 111,68g

At last, you only have to add the results: 24+14+96+11,68= 245,68g/mol. This example was for the first molecule.

See you,

5 0
2 years ago
According to the equation, 2 H2 + O2 --> 2 H20, how many moles of oxygen are required to convert 12 moles of hydrogen to wate
DerKrebs [107]

Answer:

6 moles of oxygen, or 3O2.

Explanation:

I would build a proportion for this:

\frac{2(2)}{2}  =  \frac{12}{x}  \\ 4x = 24 \\ x = 6

5 0
3 years ago
Which of these expressions are correct variations of the Combined Gas Law?
mafiozo [28]

Answer:

Both

Explanation:

The combined gas law is also known as the general gas law.

From the ideal gas law we assume that n = 1;

So;

              PV  = nRT

 and then;

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

   If we cross multiply;

                P₁V₁T₂   = P₂V₂T₁

  So;

         T₁ = T_{2} \frac{P_{1}V_{1}  }{P_{2} V_{2} }

Also;

         V₂  = V_{1} \frac{P_{1} T_{2} }{P_{2} T_{1} }

So from the choices both are correct

3 0
3 years ago
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