10 mol/h of 50 mol % NaOH is mixed with enough 6.5 mol % NaOH to produce a 10 mol

105.4 kJ

Explanation:

Given that:

10 mol/h of 50 mol % NaOH is mixed with enough 6.5 mol % NaOH to produce a 10 mol % aqueous NaOH solution

Now,if the moles of 6.5 mol% of NaOH is mixed with 10 mol of 50% NaOH

We have; Total moles = (M + 10) moles

moles of NaOH = ( 10 × 0.5) + ( M × 0.065)

The mixture then tends to be the the mole of NaOH solution

=

=

The total moles = 10 +

=

=

10% mole of NaOH & 90% moles of H₂O

Thus,    

=

=  g of NaOH

For H₂O

=

=

= g of H₂O

Finally, Total mass of mixture =

=

Heat capacity of solution = 2.8 J/g ° C

At 25 ° C, heat removed to change the temperature from 25° C to 10° C

ΔT =  ( 25 - 10 ) ° C

ΔT = 15 ° C

ΔQ = mcΔT

ΔQ = * 2.8 J/g ° C * 15° C

ΔQ = 105444 J

ΔQ = 105.4 kJ

Hence, the amount of heat to  be removed to keep the final solution at 10 ° C = 105.4 kJ

See explanation

Explanation:

Let us consider the two compounds carefully. Carbon dioxide is purely a non-polar molecular compound while the bond between the C2^2- ion and Ca^2+ is ionic thereby making CaC2 an ionic compound.

Recall that ionic compounds have high melting and boiling points due to strong ionic interactions in the structure of the compound.

On the other hand, CO2 is a non-polar substance whose molecules are held together only by weak dispersion forces.

As a result of the reasons outlined above, carbon dioxide has a low boiling point and calcium carbide has a high boiling point.