The answer is Solid.
This is on account of the substances that develop a solid are packed in a settled, firmly pressed geometric plan.
Answer:
Number of moles = 0.0005 mol.
Explanation:
Given data:
pH = 3
Volume of solution = 500 mL
Number of moles = ?
Solution:
HCl dissociate to gives H⁺ and Cl⁻
HCl → H⁺ + Cl⁻
It is known that,
pH = -log [H⁺]
3 = -log [H⁺]
[H⁺] = 10⁻³ M
[H⁺] = 0.001 M
Number of moles of HCl:
Molarity = number of moles / Volume in litter
Number of moles = Molarity × Volume in litter
Number of moles = 0.001 mol/L × 0.5 L
Number of moles = 0.0005 mol
Right now it's written in scientific notation, so you can just move the decimal place in 3.8 to the right 3 times (as it is times 10 to the third power) to get 3,800g.
H₂S
<h3>Further explanation</h3>
Given
ΔH fusion and ΔH vaporization of different substances
Required
The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes
Solution
We can use the formula :

Q=heat/energy absorbed
n = moles
The heat absorbed : 58.16 kJ
moles = 3.11
so ΔH vaporization :

The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S
(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)
Moles of PF₃ : 4
<h3>Further explanation</h3>
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Reaction

1.25 moles of P₄(s) is reacted with 6 moles of F₂(g)
Limiting reactant : the smallest ratio (mol divide by coefficient)
P₄ : F₂ =

mol PF₃ based on mol of limiting reactant(F₂), so mol PF₃ :
