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2 years ago

Help please with this tough question

Chemistry

1 answer:

Alla [95]2 years ago

5 0

Answer:

option B
option A

Explanation:

please mark me as brainliest ❤️

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What is the Oort cloud

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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.

Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams

Explanation:

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

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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)

Answer: The volume of HCl neutralized is 1.25 mL

Explanation:

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

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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

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3 years ago

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