Answer:
C
Explanation
On the reactants side there is 4 Hydrogen atoms in total and two oxygen atoms on the left however on the right there is two hydrogen atoms and one oxygen atom. Leaving this equation unbalanced
When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

Given the specific heat capacity of the solution (c) is 4.184 J/g・°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
Learn more: brainly.com/question/4400908
Answer:
Iconic bond or covalent bond
Explanation:
<h3>
<u>Answer;</u></h3>
= 930.23 mL
<h3><u>Explanation</u>;</h3>
Using the combined gas law;
P1V1/T1 = P2V2/T2
Where; P1 = 600 kPa, V1 = 800 mL, and T1 = -25 +273 = 258 K, and
V2= ?, P2 = 1000 kPa, and T2 = 227 +273 = 500 K
Thus;
V2 = P1V1T2/T1P2
= (600 ×800 ×500) / (258 × 1000)
= 930.23 mL