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3 years ago

What is the number of moles of solute in 250 mL of a 0.4 M solution?

Chemistry

1 answer:

mafiozo [28]3 years ago

3 0

Answer:

0,1 mol

Explanation:

We know that the formula of concentration is C= moles of solute/ volume

0,4 M= moles of solute/ 250 mL

Convert mL to L 250 mL =0,25 L

0,4 M x 0,25 L= moles of solute

0,1 moles= moles of solute

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4 years ago

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5) Calculate the molality of 0.210 mol of KBr dissolved in 0.075kg pure water?

Margaret [11]

Answer:

$\boxed {\boxed {\sf 2.8 \ m }}$

Explanation:

The formula for molality is:

$m=\frac{moles \ of \ solute}{kg \ of \ solvent}$

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

$moles= 0.210 \ mol \\kilograms = 0.075 \ kg$

Substitute the values into the formula.

$m= \frac{ 0.210 \ mol }{0.075 \ kg}$

Divide.

$m= 2.8 \ mol/kg= 2.8 \ m$

The molality is 2.8 moles per kilogram

5 0

3 years ago

A 12.5 g sample of granite initially at 82.0 oc is immersed into 25.0 g of water that is initially at 22.0 oc. what is the final

Paraphin [41]

The amount of heat lost by granite is equal to the amount of heat gained by water. Therefore their change in enthalpies must be equal. The opposite in sign means that one is gaining while the other is losing

ΔH granite = - ΔH water

ΔH is the change in enthalpy experienced by a closed object as it undergoes change in energy. This is expressed mathematically as,

ΔH = m Cp (T2 – T1)

Given this information, we can say that:

12.5 g * 0.790 J / g ˚C * (T2 – 82 ˚C) = - 25.0 g * 4.18 J / g ˚C * (T2 – 22 ˚C)

9.875 (T2 – 82) = 104.5 (22 – T2)

9.875 T2 – 809.75 = 2299 – 104.5 T2

114.375 T2 = 3108.75

T2 = 27.18 ˚C

The temperature of 2 objects after reaching thermal equilibrium is 27.18 ˚C.

7 0

3 years ago

a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill

baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Also,

Moles = mass (m) / Molar mass (M)

Density (d) = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRT$

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

1223.38 mmHg must be the pressure of the nitrogen gas.

5 0

3 years ago