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2 years ago

Calculate the molar concentration of 4.75 moles of NaCl in 0.50 liters of water?

Chemistry

1 answer:

worty [1.4K]2 years ago

7 0

Answer:

9.5 m

Explanation:

First, divide 4.75 m by the 0.50 l

4.75/0.50

This leaves you with 9.5 m

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What elements are in table salt

Sedaia [141]

The formula is NaCi, containing equal amounts of sodium and chloride

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3 years ago

Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin pla

RSB [31]

Answer: The enthalpy change of the reaction is -27. kJ/mol

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 25.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{25.0mL}\\\\\text{Mass of water}=(1g/mL\times 25.0mL)=25g$

To calculate the heat released by the reaction, we use the equation:

$q=mc\Delta T$

where,

q = heat released

m = Total mass = [1.25 + 25] = 26.25 g

c = heat capacity of water = 4.18 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(21.9-25.8)^oC=-3.9^oC$

Putting values in above equation, we get:

$q=26.25g\tiimes 4.18J/g^oC\times (-3.9^oC)=-427.9J=-0.428kJ$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ammonium nitrate = 1.25 g

Molar mass of ammonium nitrate = 80 g/mol

Putting values in above equation, we get:

$\text{Moles of ammonium nitrate}=\frac{1.25g}{80g/mol}=0.0156mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -0.428 kJ

n = number of moles = 0.0156 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-0.428kJ}{0.0156mol}=-27.44kJ/mol$

Hence, the enthalpy change of the reaction is -27. kJ/mol

4 0

3 years ago

2CO + O2 --> 2002

olga_2 [115]

8 moles I think I’m not sure

4 0

3 years ago

Why is it usually warmer at Earth’s equator than at its poles?

MissTica

Answer:

the equator is closer to the sun

8 0

2 years ago

What is the total number of calories of heat energy absorbed when 10 grams of water is vaporized at its normal boiling point

Darya [45]

Answer: The amount of energy absorbed by water is 5390 Calories

Explanation:

To calculate the amount of heat absorbed at normal boiling point, we use the equation:

$q=m\times L_{vap}$

where,

q = amount of heat absorbed = ?

m = mass of water = 10 grams

$L_{vap}$ = latent heat of vaporization = 539 Cal/g

Putting values in above equation, we get:

$q=10g\times 539Cal/g=5390Cal$

Hence, the amount of energy absorbed by water is 5390 Calories

6 0

3 years ago