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Jlenok [28]
3 years ago
11

What is the total number of moles of atoms in one mole of (NH4)2SO4? 1. 10 2. 11 3. 14 4. 15

Chemistry
2 answers:
anygoal [31]3 years ago
6 0

2N
8H
1S
4O
2+8+1+4=15

4.  15
Neporo4naja [7]3 years ago
5 0

Answer:

4. 15

Explanation:

The given formula is:- (NH_4)_2SO_4

1 mole of the salt contains 2 moles of nitrogen atom, 8 moles of hydrogen, 1 mole of sulfur and 4 moles of oxygen atom as can be seen from the formula.

Thus,

1 mole of salt contains total of 15 moles of the atoms which constitute the salt.

<u>Hence, 4. is the answer.</u>

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What are the main reactants and products in a neutralization reaction?
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The reactants in the neutralization reaction are an acid and a base while the products are a salt and water.

7 0
3 years ago
What happens to a gas that is enclosed in a rigid container when the temperature of the gas is increased?
yanalaym [24]
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Which of the following became a major drawback to using ice blocks for cooling food? availability expense unsanitary ice storage
son4ous [18]
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7 0
3 years ago
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A chemist must prepare 575.mL of 1.00M aqueous sodium carbonate Na2CO3 working solution. He'll do this by pouring out some 1.58M
igor_vitrenko [27]

Answer : The volume in mL of the sodium carbonate stock solution is 364 mL.

Explanation :

According to dilution law:

M_1V_1=M_2V_2

where,

M_1 = molarity of aqueous sodium carbonate

M_2 = molarity of aqueous sodium carbonate stock solution

V_1 = volume of aqueous sodium carbonate

V_2 = volume of aqueous sodium carbonate stock solution

Given:

M_1 = 1.00 M

M_2 = 1.58 M

V_1 = 575 mL

V_2 = ?

Now put all the given values in the above formula, we get:

1.00M\times 575mL=1.58M\times V_2

V_2=363.92mL\approx 364mL

Therefore, the volume in mL of the sodium carbonate stock solution is 364 mL.

8 0
2 years ago
Calculate the molar solubility of Ni(OH)2 in water. Use 2.0 * 10^-15 as the solubility product constant of Ni(OH)2.
finlep [7]
Ni(OH)₂(s) ⇄ Ni²⁺(aq) + 2OH⁻(aq)

Ksp=2.0*10⁻¹⁵

Ksp=[Ni²⁺][OH⁻]²

c=[Ni²⁺]=[OH⁻]/2

Ksp=c×(2c)²=4c³

c=∛(Ksp/4)

c=∛(2.0×10⁻¹⁵/4)=0.01995 mol/L ≈ 0.02 mol/l
8 0
2 years ago
Read 2 more answers
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