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2 years ago

What mass of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

Chemistry

1 answer:

iren [92.7K]2 years ago

3 0

91.4 grams

91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

C = mol/volume

2.45M=mol/0.5L

2.45M⋅0.5L = mol

mol = 1.225

Convert no. of moles to grams using the atomic mass of K + Cl

1.225mol * $\frac{39.1+35.5}{mol}$

mol=1.225

=1.225 mol . $\frac{74.6g}{mol}$

=1.225 . 74.6

=91.4g

therefore, 91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

What is 1 molar solution?

In order to create a 1 molar (M) solution, 1.0 Gram Molecular Weight of the chemical must be dissolved in 1 liter of water.

58.44 g make up a 1M solution of NaCl.

To learn more about molar solution visit:

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A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

musickatia [10]

Answer:

$T_2=335.42K=62.27^oC$

Explanation:

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In this case, by using the general gas law, that allows us to understand the pressure-volume-temperature relationship as shown below:

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

Thus, solving for the temperature at the end (considering absolute units of Kelvin), we obtain:

$T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{1.8L*0.75atm*(25+273.15)K}{1.2L*1.0atm} \\\\T_2=335.42K=62.27^oC$

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4 0

3 years ago

Read 2 more answers

The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl

nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1 x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0

3 years ago

At a pressure of 650 kpa 2.2L of hydrogen is used to fill a balloon to a final pressure of 115 kpa. What's the balloons volume?

Alborosie

Answer:

V₂ = 12.43 L

Explanation:

Given data:

Initial pressure = 650 KPa

Initial volume = 2.2 L

Final pressure = 115 KPa

Final volume = ?

Solution:

The given problem will be solved through the Boyles law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

650 KPa ×2.2 L = 115 KPa × V₂

V₂ = 1430 KPa. L/ 115 KPa

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3 years ago

A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

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Answer:

See explanation

Explanation:

The molecular equation shows all the compounds involved in the reaction.

The molecular equation is as follows;

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The complete ionic equation shows all the ions involved in the reaction

The complete ionic equation;

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The net Ionic equation shows the ions that actually participated in the reaction

The net ionic equation is;

2F^-(aq) + Pb^2+(aq)--------> PbF(s)

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3 years ago

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