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3 years ago

Calculate the molarity of a solution of Nach if it contains 7.2.g Nach in 100.0 mL of solution. andver: m Nach . .

Chemistry

1 answer:

Norma-Jean [14]3 years ago

5 0

Answer:

1.23 M

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = given mass of NaCl = 7.2 g

As we know , the molecular mass of NaCl = 58.5 g/mol

Moles is calculated as -

n = w / m = 7.2 g / 58.5 g/mol = 0.123 mol

Molarity is calculated as -

V = 100ml = 0.1 L (since , 1 ml = 1/1000L )

M = n / V = 0.123 mol / 0.1 L = 1.23 M

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describe in general terms an experiment to determine the molal freezing point depression constant kf of water. Assume the availa

Dvinal [7]

A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.

6 0

3 years ago

A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp

vichka [17]

Answer:

The balanced chemical equation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = $T_1= 21.36^oC$

Final temperature of the calorimeter = $T_2= 26.37^oC$

Heat absorbed by calorimeter = Q

$Q=C\times \Delta T$

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

$\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g$

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

$\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol$

3 0

3 years ago

Which of the following is not the same as 2.97 milligrams?

Len [333]

Hi There!!!!!!

Which of the following is not the same as 2.97 milligrams?A)0.297 cg
B)0.00297 g
C) 0.0000297 KG

Answer:0.00297 g

5 0

2 years ago

Read 2 more answers

(06.03 MC) A 50.0 mL sample of gas at 20.0 atm of pressure is compressed to 40.0 atm of pressure at constant temperature. What i

oee [108]

Answer:

New volume is 25.0 mL

Explanation:

Let's assume the gas sample behaves ideally.

According to combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where, $P_{1}$ and $P_{2}$ represent initial and final pressure respectively

$V_{1}$ and $V_{2}$ represent initial and final volume respectively

$T_{1}$ and $T_{2}$ represent initial and final temperature (in kelvin) respectively

Here, $T_{1}=T_{2}$ , $V_{1}=50.0mL$ , $P_{1}=20.0atm$ and $P_{2}=40.0atm$

So, $V_{2}=\frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}=\frac{(20.0atm)\times (50.0mL)}{40.0mL}=25.0mL$

So, the new volume is 25.0 mL

6 0

3 years ago

A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of

ICE Princess25 [194]

Answer:

The gas obeys Boyle’s law and the value of $k_i\&k_f$ both are equal to 40.0 atm L.

Explanation:

Initial volume of the gas = $V_1=10.0 L$

Initial pressure of the gas = $P_1=4.0 atm$

Final volume of the gas = $V_2=5.0 L$

Final pressure of the gas = $P_2=8.0 atm$

This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

$PV=k$

The equation given by this law is:

$P_1V_1=P_2V_2$

$P_1\propto \frac{1}{V_1}$

$P_1V_1=k_i$

$k_i=4.0 atm\times 10.0 L = 40.0 atm L$

$P_2\propto \frac{1}{V_2}$

$P_V_2=k_f$

$k_f=8.0 atm\times 5.0 L = 40.0 atm L$

$k_i=k_f=40.0 atm L$

The gas in the cylinder is obeying Boyle's law.

The gas obeys Boyle’s law and the value of $k_i\&k_f$ both are equal to 40.0 atm L.

6 0

3 years ago