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2 years ago

1. How many grams of potassium chloride (KCI) can be dissolved at 10°C?

Chemistry

1 answer:

Allisa [31]2 years ago

3 0

Answer:

At 10 ∘C , the solubility of potassium chloride is measured to be 31.2⋅g per 100⋅g of water.

Explanation:

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Name the organic compound of ch3chohch2ch3

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Answer:

Flavouring agent 2-Butanol, or sec-butanol, is an organic compound with formula CH3CH(OH)CH2CH3.

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2 years ago

Calculate the concentration of CO2 in water at 25 degrees Celsius when the pressure of CO2 over the solution is 4.5 atm. At 25 d

OLEGan [10]

Answer : The concentration of $CO_2$ is, 0.12 M

Explanation :

Using Henry's law :

$C_{CO_2}=k_H\times p_{CO_2}$

where,

$C_{CO_2}$ = concentration of $CO_2$ = ?

$p_{CO_2}$ = partial pressure of $CO_2$ = 4.5 atm

$k_H$ = Henry's law constant = $3.1\times 10^{-2}mol/L.atm$

Now put all the given values in the above formula, we get:

$C_{CO_2}=3.1\times 10^{-2}mol/L.atm\times (4.5atm)$

$C_{CO_2}=0.1395M\approx 0.12M$

Thus, the concentration of $CO_2$ is, 0.12 M

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3 years ago

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3 years ago

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Explanation:

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3 years ago

Read 2 more answers

In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water shoul

BARSIC [14]

<u>Answer:</u> The mass of water that should be added in 203.07 grams

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

m = molality of barium iodide solution = 0.175 m

$m_{solute}$ = Given mass of solute (barium iodide) = 13.9 g

$M_{solute}$ = Molar mass of solute (barium iodide) = 391.14 g/mol

$W_{solvent}$ = Mass of solvent (water) = ? g

Putting values in above equation, we get:

$0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g$

Hence, the mass of water that should be added in 203.07 grams

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3 years ago