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Margaret [11]
3 years ago
9

What is the predicted order of first ionization energies from highest to lowest for aluminum, phosphorus, silicon, and sulfur?

Chemistry
2 answers:
DochEvi [55]3 years ago
7 0

This is the answer


P > S > Si > Al


Phosphorus is in group 15 and sulfur is in group 16 so the ionization energies is a little bit lower the phosphorus.

e-lub [12.9K]3 years ago
5 0

the correct answer is D . (P > S > Si > Al)

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The solubility of CO2 in water is 0.161 g/100 mL at 20oC and a partial pressure of CO2 of 760 mmHg. What partial pressure of CO2
Schach [20]

<u>Answer:</u> The partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

<u>Explanation:</u>

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

The equation given by Henry's law is:

C_{CO_2}=K_H\times p_{CO_2}       ......(1)

where,

C_{CO_2 = solubility of carbon dioxide in water = 0.161 g/100 mL

K_H = Henry's constant = ?

p_{CO_2} = partial pressure of carbon dioxide = 760 mmHg

Putting values in equation 1, we get:

760mmHg=K_H\times 0.161g/100mL\\\\K_H=\frac{760mmHg}{0.161g/100mL}=4720.5g.mmHg/100mL

Now, calculating the pressure of carbon dioxide using equation 1, we get:

C_{CO_2 = solubility of carbon dioxide in water = 0.886 g/100 mL

K_H = Henry's constant = 4720.5 g.mmHg/100 mL

p_{CO_2} = partial pressure of carbon dioxide = ?

Putting values in equation 1, we get:

p_{CO_2}=4720.5g.mmHg/100mL\times 0.886g/100mL=4182.4mmHg

Hence, the partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

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A welding torch produces a flame by burning acetylene fuel in the presence of oxygen the energy released from the acetylene fuel
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Answer:

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3 years ago
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A 5.000 g sample of Niso4 H2O decomposed to give 2.755 g of anhydrous NiSO4.
Vinvika [58]

Answer:

a) 7.0.

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d) 44.9%.

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<u><em>a) What is the formula of the hydrate?</em></u>

The mass of the hydrated sample (NiSO₄.xH₂O) = 5.0 g,

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The mass of water = 5.0 g - 2.755 g = 2.245 g.

∴ no. of moles of water = mass/molar mass = (2.245 g)/(18.0 g/mol) = 0.1247 mol.

∴ no. of moles of anhydrous salt (NiSO₄) = mass/molar mass = (2.755 g)/(154.75 g/mol) = 0.0178 mol.

∴ water of crystallization in the sample (x) = no. of moles of water/no. of moles of anhydrous salt (NiSO₄) = (0.1247 mol)/(0.0178 mol) = 7.0.

<u><em>b) What is the full chemical name for the hydrate?</em></u>

The name of the salt (NiSO₄.7H₂O) is Nickel sulfate hepta hydrate.

<u><em>c) What is the molar mass of the hydrate? </em></u>

(NiSO₄.7H₂O)

The molar mass = molar mass of NiSO₄ + 7(molar mass of H₂O) = (154.75 g/mol) + 7(18.0 g/mol) = 280.83 g/mol.

<em><u>d) What is the mass % of water in the hydrate?</u></em>

The mass % of water = (mass of water)/(mass of hydrated sample) x 100 = (2.245 g)/(5.0 g) x 100 = 44.9%.

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