Answer:
12.9 m³ is the new volume
Explanation:
As the temperature keeps on constant, and the moles of the gas remains constant too, if we decrease the pressure, the volume will increase. If the volume is decreased, pressure will be higher.
The relation is this: P₁ . V₁ = P₂ . V₂
1 atm . 0.93m³ = 0.072 atm . V₂
0.93m³ .atm / 0.072 atm = V₂
V₂ = 12.9 m³
In conclusion and as we said, pressure has highly decreased so volume has highly increased.
Answer:
300 mL
Explanation:
the unit formula of calcium phosphate is Ca3(PO4)2
molar mass of Ca3(PO4)2 = (3×40 + 2×31 + 8×16) g/mol = 310 g/mol
n = m/M = 35 g/(310 g/mol)
c = n/V
V = n/c = [35 g/(310 g/mol)]/0.375 mol/L
V = 0.30 L = 300 mL
Answer:

Explanation:
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 28.01 17.03
N₂ + 3H₂ ⟶ 2NH₃
m/g: 240.0
(a) Moles of NH₃

(b) Moles of N₂

(c) Mass of N₂

Oxidation of D -Ribose in presence of hypobromous acid gives D-Ribonic acid