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melisa1 [442]
3 years ago
5

An aqueous solution contains 32.7% kcl (weight/weight %). how many grams of water are contained in 100 g of this solution

Chemistry
1 answer:
igor_vitrenko [27]3 years ago
5 0
Weight/weight percentage is the percentage of mass of solute from the mass of the whole solution
w/w % = 32.7 %
this means that in a solution of 100 g - mass of KCl is 32.7 g 
since solution is made of water and KCl
then the mass of water is - 100 - 32.7 = 67.3 g
therefore in 100 g of solution - 67.3 g of the solution is water
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A gas has a volume of 3.25 liters at 54 C and 231 kPa of pressure. At what temperature will the same gas take up 4.35 liters of
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Answer: 318 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 231 kPa

P_2 = final pressure of gas = 168 kPa

V_1 = initial volume of gas = 3.25 L

V_2 = final volume of gas = 4.35 L

T_1 = initial temperature of gas = 54^oC=273+54=327K

T_2 = final temperature of gas = ?

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

\frac{231\times 3.25}{327}=\frac{168\times 4.35}{T_2}

T_2=318K

At 318 K of temperature will the same gas take up 4.35 liters of space and have a pressure of 168 kPa

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3 years ago
Helium gas in a cylinder is under 1.12atm pressure at 25.0C. What will be the pressure if the temperature increases to 37.0C?
Irina18 [472]

Answer:

p_2=1.17atm

Explanation:

Hello!

In this case, considering that the Gay-Lussac's law allows us to relate the temperature-pressure problems as directly proportional relationships:

\frac{p_2}{T_2} =\frac{p_1}{T_1} \\\\

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p_2=\frac{p_1T_2}{T_1} \\\\p_2=\frac{1.12atm*310.15K}{298.15K}\\\\p_2=1.17atm

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irina1246 [14]

If It takes 60mL of 0.20M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H_2CO_3), then the concentration of the carbonic acid is 0.24M

The reaction between NaOH solution and H_2CO_3 is written below

2NaOH + H_2CO_3 \rightarrow Na_2CO_3 + 2H_2O

Volume of NaOH, V_B = 60 ml

Volume of H_2CO_3, V_A=25 ml

Molarity of H_2CO_3, C_A=?

Molarity of NaOH, C_B=0.20M

Number of moles of H_2CO_3, n_A=1

Number of moles of NaOH, n_B=2

The mathematical equation for neutralization reaction is:

\frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B}

Substitute  C_B=0.2 M,  n_A=1,  n_B=2,  V_B = 60ml, and  V_A=25 ml into the equation above in order to solve for C_A

\frac{C_A \times 25}{0.2 \times 60}=\frac{1}{2}  \\\\50C_A=12\\\\C_B=\frac{12}{50} \\\\C_B=0.24M

Therefore, the concentration of the carbonic acid is 0.24M

Learn more here: brainly.com/question/25943090

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2 years ago
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