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Rufina [12.5K]
3 years ago
14

At a certain temperature, the solubility of N2 gas in water at 4.07 atm is 95.7 mg of N2 gas/100 g water . Calculate the solubil

ity of N2 gas in water, at the same temperature, if the partial pressure of N2 gas over the solution is increased from 4.07 atm to 10.0 atm .
Chemistry
1 answer:
sertanlavr [38]3 years ago
6 0

Answer: Thus the solubility of N_2 gas in water, at the same temperature, if the partial pressure of gas is 10.0 atm is 235mg/100g.

Explanation:-

The Solubility of N_{2} in water can be calculated by Henry’s Law. Henry’s law gives the relation between gas pressure and the concentration of dissolved gas.

Formula of Henry’s law,  C=k_{H}P.

k_{H}= Henry’s law constant = ?

The partial pressure (P) of N_{2} in water = 4.07 atm

\C= k_{H}\times P\\95.7mg=k_{H}\times 4.07

k_{H}=23.5

At pressure of 10.0 atm

C= k_{H}\times P\\C=23.5\times 10.0=235mg/100mg

Thus the solubility of N_2 gas in water, at the same temperature, is 235mg/100g

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<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

Mg(NO_3)_2(aq.)+2NH_4OH(aq.)\rightarrow Mg(OH)_2(s)+2NH_4NO_3(aq.)

A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Mg^{2+}(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)

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How many chloride ions are in 2.6 moles of CaCl2?
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Answer:

31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .

Explanation:

Given data:

Number of moles of CaCl₂ = 2.6 mol

Number of Cl₂ ions = ?

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CaCl₂  → Ca²⁺ + 2Cl⁻

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

In one mole of CaCl₂ there are two moles of chloride ions present.

In 2.6 mol:

2.6×2 = 5.2 moles

1 mole Cl⁻ =   6.022 × 10²³ number of Cl⁻ ions

5.2 mol ×  6.022 × 10²³ number of Cl⁻  / 1mol

31.31× 10²³ number of Cl⁻

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Answer:

Rubidium-85=61.2

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Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

<u>Rubidium-85 </u>

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

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Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

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Rubidium-85=61.2

<u>Rubidium-87</u>

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

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Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

  • mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

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Add the two numbers together.

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