Arsenic has one stable isotope:75/33 As.

What mass of solid that has a molar mass 89.0 g/mol should be added to 100.0 g of benzene to raise the boiling point of benzene

taurus [48]

On the basis of given data:

Mass of solvent (benzene) = 100 gm or 0.1 Kg

Molar mass of solute (solid) = 89 g/mol

Increase in boiling point (deltaTb) = 2.42 degree C

The boiling point elevation constant k = 2.53 C.kg/mol

There is a need to determine the mass of the solid to be added, the elevation in boiling point is proportional to the m (molality) of the solute;

ΔTb = k.m

Here, m that is molality = moles of solute/kg of solvent

Therefore,

ΔTb = k (benzene) × moles of solid/kg of benzene

2.42 = 2.53 × moles of solid/0.1

moles of solid = 0.0956 moles

The molar mass of solid = 89 g/mol

Thus, the mass of solid = 0.0956 moles × 89 g/mol

= 8.508 g

Thus, the mass of solid to be added is 8.508 g.

3 0

3 years ago

Which of the following best describes the forces that generate solar power

Sergeeva-Olga [200]

I believe the answer is A (it was on another Brainly question as well).

8 0

3 years ago

What would ne the acceleration in a body moving wit uniform velocity and why

meriva

Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.

Hope this helps :) :)

3 0

3 years ago

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Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

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

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

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

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

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

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

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

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$