The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

4, because there are 4 quarts in 1 gallon.
Another way to think about it is 1 gallon= $1 and 1 quart= 25 cents
There are 4 quarters in a dollar.
Hope this helped..
Answer:

Explanation:
We want to convert from moles to grams, so we must use the molar mass.
<h3>1. Molar Mass</h3>
The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.
- Phosphorus (P): 30.973762 g/mol
- Iodine (I): 126.9045 g/mol
Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.
- I₃: 126.9045 * 3=380.7135 g/mol
- PI₃: 30.973762 + 380.7135 = 411.687262 g/mol
<h3>2. Convert Moles to Grams</h3>
Use the molar mass as a ratio.

We want to convert 3.14 moles to grams, so we multiply by that value.

The units of moles of PI₃ cancel.


<h3>3. Round</h3>
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.
The 2 in the ones place tells us to leave the 9.

3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>
Moles of K = 32.4/39 = 0.83 mole. According to stoichiometry, 2 moles of K produces 1mole of H2. Therefore, 0.83 mole of K produces = 0.83/2 = 0.415 moles of H2. Therefore number molecules of H2 = moles of H2 x 6.02 x 10^23 = 2.4983 x 10^23 molecules. Hope this helps!