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Juliette [100K]
2 years ago
6

Help please with this tough question

Chemistry
1 answer:
Alla [95]2 years ago
5 0

Answer:

  1. option B
  2. option A

Explanation:

please mark me as brainliest ❤️

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The Oort cloud sometimes called the Öpik–Oort cloud it was first described in 1950 by Dutch astronomer Jan Oort
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3 years ago
At the Henry's Law constant for carbon dioxide gas in water is . Calculate the mass in grams of gas that can be dissolved in of
Dvinal [7]

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of carbon dioxide that can be dissolved is 1.7 grams

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{CO_2}=K_H\times p_{CO_2}

where,

K_H = Henry's constant = 0.031M/atm

C_{CO_2} = molar solubility of carbon dioxide gas

p_{CO_2} = partial pressure of carbon dioxide gas = 2.92 atm

Putting values in above equation, we get:

C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of carbon dioxide = ? g

Molar mass of carbon dioxide = 44 g/mol

Molarity of solution = 0.0905mol/L

Volume of solution = 425 mL

Putting values in above equation, we get:

0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g

Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams

8 0
3 years ago
16 meters per hour to miles per sec
ss7ja [257]

Answer:

2.76 (full answer) 2.7617e-6

Explanation:

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2 years ago
What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa
almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of stomach acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL

Hence, the volume of HCl neutralized is 1.25 mL

3 0
3 years ago
Interesting facts about coal
babunello [35]
Coal will eventually run out as it is an non-renewable energy. these are when they cannot be reproduced to make more,unlike would as u can replant them to create trees
8 0
3 years ago
Read 2 more answers
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