How myny moles are present in 5.6x1022 atoms of Ne? (0.09 moles Ne) Sot ne10.05

A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

ladessa [460]

Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Let's call chloroform C and acetone A.

Molar concentration of C = Moles of C/Litres of solution

(a) Moles of C

Assume 0.187 mol of C.

That takes care of that.

(b) Litres of solution

Then we have 0.813 mol of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

If there is no change of volume on mixing.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration = moles of solute/kilograms of solvent

Moles of C = 0.187 mol

Mass of A = 47.22 g = 0.047 22 kg

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

4 0

2 years ago

How many moles of O2 are produced when 0.500mol of KO2 reacts in this fashion?

Shtirlitz [24]

I think you means the KO2 reacts with H2O. The equation of this reaction is 4KO2+2H2O->4KOH +3O2. The ratio of mole number of O2 and KO2 is 3:4. So the mole number of O2 produced is 0.500/4*3=0.375 mol.

6 0

2 years ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

7 0

2 years ago

Read 2 more answers

If 16.0 mL of acetone is dissolved in water to make 155 mL of solution what is the concentration expressed in volume/volume % of

Lady_Fox [76]

Percentage by volume of solution is the percentage volume of solute in total volume of solution.
Volume percentage (v/v%) = volume of solute / total volume of solution x 100%
volume of solute - 16.0 mL
total volume of solution - 155 mL
v/v% = 16.0 / 155 x 100% = 10.32%
this means that in a volume of 100 mL solution, 10.32 mL is acetone.

7 0

2 years ago

the air we breathe is approximately 21% oxygen.atypical breath has a volume of 450 ml.how many grams of O2 are in a breath of ai

nirvana33 [79]

We breathe around an average of 500 ml of air in 1 normal (or relax) breathing. The lungs already have around 3 litres of reserve breathing. Since there is around 21% oxygen in air, we breathe in around 105 ml of Oxygen from the air. That is equal to around 0.14 g of Oxygen.

6 0

2 years ago