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Vedmedyk [2.9K]
2 years ago
12

Find the normality of 0.321 g sodium carbonate in a 250 mL solution.

Chemistry
1 answer:
Radda [10]2 years ago
5 0

Answer:

the normality of the given solution is 0.0755 N

Explanation:

The computation of the normality of the given solution is shown below:

Here we have to realize the two sodiums ions per carbonate ion i.e.

N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)

= 0.1886eq ÷ 0.2500L

= 0.0755 N

Hence, the normality of the given solution is 0.0755 N

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Andre45 [30]

Answer:

The fertilizer ammonium sulfate, (NH4)2SO4, is prepared by the reaction of ammonia, NH3, with sulfuric acid: 2 NH3 (g) + H2SO4 (aq)  (NH4)2SO4 (aq)

How many moles of ammonium sulfate are produced if 1.50 moles of ammonia react completely at STP? Answer: The balanced equation indicates that 2 moles of NH3 react to produce 1 mole of (NH4)2SO4.

1.50 moles NH3 x 1 mol (NH4)2SO4 = 0.750 moles (NH4)2SO4

Explanation:

6 0
3 years ago
Which statement below correctly describes the relationship between Q and K for both reactions? Are these reactions spontaneous a
nikklg [1K]

Answer:

Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Explanation:

Hello,

In this case, the undergoing chemical reactions with their proper Gibbs free energy of reaction are:

A->B;\Delta _rG^o=-13 kJ/mol

C ->D ;\Delta _rG^o=3.5 kJ/mol

The cellular concentrations are as follows: [A] = 0.050 mM, [B] = 4.0 mM, [C] = 0.060 mM and [D] = 0.010 mM.

For each case, the reaction quotient is:

Q_1=\frac{4.0mM}{0.050mM}=80\\ Q_2=\frac{0.010mM}{0.060mM}=0.167

A typical temperature at a cell is about 30°C, in such a way, the equilibrium constants are:

K_1=exp(-\frac{-13000J/mol}{8.314J/mol*K*303.15K} )=173.8\\K_2=exp(-\frac{3500J/mol}{8.314J/mol*K*303.15K} )=0.249

Therefore, Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Best regards.

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