The solute is the part of the solution that dissolves in the second component (usually a fluid). Therefore, for the mentioned solution, the solute is ehyl alcohol since it is the one dissolving in water.
As for the solvent, it is the component in which the solute dissolves. In this case, it is water.
The ice sculpture would take longer to melt than the ice cube
Answer:
12.07 g.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>2NH₃(g) + CO₂(g) → H₂NCONH₂(g) + H₂O(g),</em>
It is clear that 2.0 moles of NH₃ react with 1.0 mole of CO₂ to produce 1.0 mole of H₂NCONH₂ and 1.0 moles of H₂O.
- Consider the reaction proceeds at STP conditions:
At STP, 9.0 L of NH₃ react with an excess of CO₂ gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of NH₃ represents → 22.4 L.
??? mol of NH₃ represents → 9.0 L.
∴ 9.0 L of NH₃ represents = (1.0 mol)(9.0 L)/(22.4 L) = 0.4018 mol.
- To find the no. of moles of urea (H₂NCONH₂) produced:
<u><em>Using cross multiplication:</em></u>
2.0 mol of NH₃ produce → 1.0 mol of H₂NCONH₂, from stichiometry.
0.4018 mol of NH₃ produce → ??? mol of H₂NCONH₂.
∴ The no. of moles of H₂NCONH₂ = (1.0 mol)(0.4018 mol)/(2.0 mol) = 0.201 mol.
- Now, we can find the mass of H₂NCONH₂ produced:
<em>mass = n * molar mass</em> = (0.201 mol) * (60.06 g/mol) = <em>12.07 g.</em>
Answer: The standard enthalpy change is -607kJ
Explanation:
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{C_6H_6}\times \Delta H_f^0_{(C_6H_6)}]-[n_{C_2H_2\times \Delta H_f^0_{(C_2H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BC_6H_6%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_6H_6%29%7D%5D-%5Bn_%7BC_2H_2%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_2H_2%29%7D%5D)
where,
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times 83)]-[(3\times 230)]=-607kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%2083%29%5D-%5B%283%5Ctimes%20230%29%5D%3D-607kJ)
The standard enthalpy change is -607kJ
Answer:
what are the options? I'll edit my answer to the correct one once you inform me lol