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Usimov [2.4K]
2 years ago
14

If the solubility of a gas in water at 25°C is 22.25 mg/L at 2.5 atm of pressure, what is its solubility (in mg/L) at 25°C

Chemistry
1 answer:
tamaranim1 [39]2 years ago
8 0

Answer:

8.9 mg/l

Explanation: Temp doesnt matter so throw that out automatically then your equation is;

S1/P1=S2/P2

We are looking for S2 and that equation is;

S2=S1*P2/P1 and that is S2=22.25*1/2.5

A little bit of simple math and you get your answer: 8.9 mg/l

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Chemist and physicist marie curie became famous for her pioneering research on radioactivity. What are the two chemical elements
wolverine [178]

Answer:

The two elements are POLONIUM and RADIUM.

Explanation:

Maria Curie is a French physicist and chemist, though she was of a Polish naturals. She was the first woman to receive a Noble Price which she earned for conducting leading and head way research on radioactivity. She discovered the theory of radioactivity; also the techniques isolating radioactive isotopes. These helped her and her husband discover Polium and Radium.

4 0
3 years ago
Read 2 more answers
Use the ideal gas law to calculate the concentrations of nitrogen and oxygen present in air at a pressure of 1.0 atm and a tempe
notka56 [123]

Answer:

[N₂] = 0.032 M

[O₂] = 0.0086 M

Explanation:

Ideal Gas Law → P . V = n .  R . T

We assume that the mixture of air occupies a volume of 1 L

78% N₂ → Mole fraction of N₂ = 0.78

21% O₂  → Mole fraction of O₂ = 0.21

1% another gases  → Mole fraction of another gases = 0.01

In a mixture, the total pressure of the system refers to total moles of the mixture

1 atm . 1L = n . 0.082L.atm/mol.K . 298K

n = 1 L.atm / 0.082L.atm/mol.K . 298K → 0.0409 moles

We apply the mole fraction to determine the moles

N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂

O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂

4 0
3 years ago
Three common gaseous compounds of nitrogen and oxygen of different elementary composition are known: (A) laughing gas containing
Ede4ka [16]

Answer:

Please find how these data prove the law of multiple proportions below

Explanation:

The law of multiple proportions was proposed by an English chemist called John Dalton. The law states that when two elements combine and to form more than one compound. The weights/masses of the second element in the two compounds, which combines with a fixed ratio of the first element, is in a simple whole number ratio.

In this question, Nitrogen is said to combine with oxygen to give three different compounds as follows:

A) laughing gas containing 63.65% nitrogen i.e. 0.6365g

This means that the mass of oxygen will be (1-0.6365) = 0.3635g

B) colorless gas containing 46.68% nitrogen i.e. 0.4668g

This means that the mass of oxygen will be 0.5332g

C) brown toxic gas containing 30.45% nitrogen i.e. 0.3045g

This means that the mass of oxygen will be 0.6955g

The ratios of oxygen in the three compounds is therefore:

0.3635: 0.5332: 0.6955

Divide this ratio by the smallest number (0.3635)

0.3635/0.3635 = 1

0.5332/0.3635 = 1.467

0.6955/0.3635 = 1.913

Multiply this ratio by 2, we have:

2: 2.9 : 3.8

Hence, the simple whole number ratio is 2:3:4.

This proves the law of multiple proportions that oxygen is in simple whole number ratio in the three different compounds.

4 0
3 years ago
The heat of reaction for the combustion of propane is –2,045 kJ. This reaction is: C3H8(g) +502 (g) 3CO2 (g) + 4H2O (g). Determi
Ede4ka [16]

Answer:

\Delta _fH_{C_3H_8}=-102.7kJ/mol

Explanation:

Hello,

In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

\Delta _cH=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-5*\Delta _fH_{O_2}

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

\Delta _fH_{C_3H_8}=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-5*\Delta _fH_{O_2}-\Delta _cH\\\\\Delta _fH_{C_3H_8}=3*(-393.5)+4*(-241.8)-5*0-(-2045)\\\\\Delta _fH_{C_3H_8}=-102.7kJ/mol

Best regards.

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3 years ago
a chemist uses hot hydrogen gas to convert chromium (iii) oxide to pure chromium. how many grams of hydrogen are needed to produ
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Answer:

Explanation:

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