All categories

3 years ago

A solution is made containing 4.92 g of sodium lithium chloride per 347 g of water.

Chemistry

1 answer:

Margaret [11]3 years ago

7 0

% mass of a solution is the mass of the solute present per 100 g of solution. It can be calculated using the formula,

$Mass % = \frac{Mass of solute (g)}{Mass of solution (g)} * 100$

Mass of solute = 4.92 g

Mass of solvent, water = 347 g

Total mass of solution = 347 g + 4.92 g = 351.92 g

Mass % of the solute = $\frac{4.92 g}{351.92 g} * 100 = 1.398 %$

Therefore, the weight/weight % = 1.398 %

You might be interested in

Radio station broadcast signals two different frequency bands. These are Called ?

rusak2 [61]

Answer:

I think it is AM and frequency

Explanation:

Sorry if i'm wrong ;)

6 0

3 years ago

Read 2 more answers

Which of the following ions/atoms has the smallest radius?

Anna11 [10]

Answer:

B I believe im sorry if it's not right

8 0

3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

Differentiate between subjective and objective mineral properties

malfutka [58]

Subjective minerals are minerals that are subjected to so something specific. Objective minerals are minerals that have an object they are made for.
although this is a small difference, in the scientific world with out this you could go totally wrong

5 0

3 years ago

Which of the following statements is true?

pashok25 [27]

The answer is longitudinal waves more particles at right angles

3 0

4 years ago