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kicyunya [14]
3 years ago
9

A solution contains 2 meq/ml of kcl. if a 20 ml ampule of this solution is diluted to 1 liter what is the percent strength of th

e resulting solution
Chemistry
1 answer:
Serjik [45]3 years ago
7 0

We are given that the concentration of the KCl is 2 meq / mL. Assuming that the ampule also has exactly this concentration, therefore:

 

amount of KCl in ampule = (2 meq / mL) * (20 mL)

amount of KCl in ampule = 40 meq

 

This amount of KCl is now inside a solution of 1 Liter (also equivalent to 1000 mL), therefore the new concentration in the resulting solution is:

new concentration = 40 meq / 1000 mL

new concentration = 0.04 meq / mL

 

Since 0.04 in decimal is 4% in percentage, therefore the strength of the resulting solution is 4% KCl.

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20.1 g of aluminum and 219 g of chlorine gas react until all of the aluminum metal has been converted to AlCl3. The balanced equ
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Answer:

<em>The amount of Cl2 gas left , after the reaction goes to completion is : </em><u><em>139.655 grams</em></u>

Explanation:

Molar mass : It is the mass in grams present in one mole of the substance.

Moles of the substance is calculated by:

Moles=\frac{Mass}{Molar\ mass}

2Al(s)+3Cl_{2}(g)\leftarrow 2AlCl_{3}(g)

According  to this equation:

2 mole of Al = 3 mole of Cl2 = 2 mole of AlCl3

Molar mass of Al = 27.0 g/mol

Mass of Al = 20.1 gram

Moles of Al present in the reaction :

Moles=\frac{Mass}{Molar\ mass}

Moles=\frac{20.1}{26.98}

Moles of Al = 0.744

Similarly calculate the moles of Cl2

Molar mass of Cl2 = 71.0 g/mol

Mass = 219 gram

Moles=\frac{Mass}{Molar\ mass}

Moles=\frac{219}{70.98}

Moles of Cl2 = 3.08 moles

According to equation,

2 mole of Al reacts with = 3 mole of Cl2

1 moles of Al reacts with = 3/2  mole of Cl2

0.744 moles of Al reacts with = 3/2(0.744) moles of Cl2

= 1.116 moles of Cl2

But actually present Cl2 = 3.08 moles

Hence Al is the limiting reagent , and Cl2 is the excess reagent.

The whole Aluminium Al get consumed during the reaction.

The amount of Cl2 in excess = Total Cl2 - Cl2 consumed

Cl2 in excess = 3.08 - 1.116 = 1.964 moles

<u>Cl2 in grams</u><u> </u>= 1.964 x 70.90 <u>= 139.655 grams</u>

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Despite being the only metal that is liquid at room temperature, mercury has the smallest liquid range of any metal. It becomes a solid at -38.83°C and a gas at 356.7°C.
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A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure su
Tom [10]

Answer:

V_2 = 4.87 * 10^3

Explanation:

This question is an illustration of ideal Gas Law;

The given parameters are as follows;

Initial Temperature = 25C

Initial Volume = 4.5 * 10³L

Required

Calculate the volume when temperature is 50C

NB: Pressure remains constant;

Ideal Gas Law states that;

PV = nRT

The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n

Divide both sides by PT

\frac{PV}{PT} = \frac{nRT}{PT}

\frac{V}{T} = \frac{nR}{P}

Represent \frac{nR}{P} with k

\frac{V}{T} = k

k = \frac{V_1}{T_1} = \frac{V_2}{T_2}

At this point, we can solve for the required parameter using the following;

\frac{V_1}{T_1} = \frac{V_2}{T_2}

Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;

From the given parameters;

V1 = 4.5 * 10³L

T1 = 25C

T2 = 50C

Convert temperatures to degree kelvin

V1 = 4.5 * 10³L

T1 = 25 +273 = 298K

T2 = 50 + 273 = 323K

Substitute values for V1, T1 and T2 in \frac{V_1}{T_1} = \frac{V_2}{T_2}

\frac{4.5 * 10^3}{298} = \frac{V_2}{323}

Multiply both sides by 323

323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323

323 * \frac{4.5 * 10^3}{298} = V_2

V_2 = 323 * \frac{4.5 * 10^3}{298}

V_2 = \frac{323 * 4.5 * 10^3}{298}

V_2 = \frac{1453.5 * 10^3}{298}

V_2 = 4.87 * 10^3

Hence, the final volume at 50C is V_2 = 4.87 * 10^3

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3 years ago
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