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WINSTONCH [101]

1 year ago

13

If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the solution is 275 mL, what will

the molarity of the solution be?

2 answers:

tatiyna1 year ago

8 0

Answer:

Explanation:

Molarity = number of moles / volume

If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the solution is 275 mL, which is half of 550 mL, the molarity of the solution with the same number of moles of KCl is 3.5 * 2 = 7.00 M

Likurg_2 [28]1 year ago

4 0

C1=3.5M
C2=?
V1=550mL
V2=275mL

$\\ \rm\rightarrowtail C1V1=C2V2$

$\\ \rm\rightarrowtail 3.5(550)=275C2$

$\\ \rm\rightarrowtail C2=3.5(2)$

$\\ \rm\rightarrowtail C2=7M$

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

vladimir1956 [14]

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ $\frac{5.60}{2}gallons = 2.80 gallons$

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = $(M__1}) = Density*Volume$

$= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 11659.06grams$

On the other hand; the mass of water = $(M__2})= Density*Volume$

$= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 10577.95 grams$

Molarity = $\frac{massof solute*1000}{molarmassof solute*massofwater}$

= $\frac{11659.06*1000}{62.07*10577.95}$

= 17.757 m

≅ 17.76 m

∴ the boiling point of the solution is calculated using the boiling‑point elevation constant for water and the Molarity.

$\Delta T_{boiling} = k_{boiling}M$

where,

$k_{boiling}$ = 0.512 °C/m

$\Delta T_{boiling}$ = 100°C + 17.56 × 0.512

= 109.09 °C

6 0

3 years ago

Fill in the following information for the element

e-lub [12.9K]

Answer:

Period: 6 block

Group: f block

Family: Lanthanides

5 0

3 years ago

Read 2 more answers

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

A

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

C

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

3 years ago

A 0.1 mm sample of human blood has approximately 6000 red blood cells. An adult typically has 5.0 L of blood. How many red blood

olga55 [171]

3.0 × 10¹¹ RBC's (or) 3E11 RBC's

Solution:

Step 1: Convert mm³ into L;

As,

1 mm³ = 1.0 × 10⁻⁶ Liters

So,

0.1 mm³ = X Liters

Solving for X,

X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³

X = 1.0 × 10⁻⁷ Liters

Step 2: Calculate No. of RBC's in 5 Liter Blood:

As given

1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's

So,

5.0 Liters of Blood will contain = X RBC's

Solving for X,

X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters

X = 3.0 × 10¹¹ RBC's

Or,

X = 3E11 RBC's

5 0

3 years ago

How would you prepare 3.5 L of a 0.9M solution of KCl?

Fudgin [204]

Calculate the mass of the solute <span>in the solution :

Molar mass KCl = </span><span>74.55 g/mol

m = Molarity * molar mass * volume

m = 0.9 * 74.55 * 3.5

m = 234.8325 g

</span><span>To prepare 0.9 M KCl solution, weigh 234.8325 g of salt in an analytical balance, dissolve in a beaker, shortly after transfer with the help of a funnel of transfer to a volumetric flask of 100 cm</span>³<span> and complete with water up to the mark, then cover the balloon and finally shake the solution to mix

hope this helps!</span>

8 0

3 years ago