First row: HCl, ZnCl2, FeCl3, AlCl3, BaCl2, PbCl4
Second row: H3P, Zn3P2, FeP, AlP, Ba3P2, Pb3P4
Third row: HNO3, Zn(NO3)2, Fe(NO3)3, Al(NO3)3, Ba(NO3)2, Pb(NO3)4
Fourth row: ZnO, Fe2O3, Al2O3, BaO, PbO2
Fifth row: HCaF2, Zn(CaF2)2, Fe(CaF2)3, Al(CaF2)3, Ba(CaF2)2, Pb(CaF2)4
Sixth row: H2SO4, ZnSO4, Fe2(SO4)3, Al2(SO4)3, BaSO4, Pb(SO4)2
Answer:
131 atm
Explanation:
To find the new pressure, you need to use Boyle's Law:
P₁V₁ = P₂V₂
In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new pressure (P₂) by plugging the given values into equation and simplifying.
P₁ = 3.88 atm P₂ = ? atm
V₁ = 7.74 L V₂ = 0.23 L
P₁V₁ = P₂V₂ <----- Boyle's Law
(3.88 atm)(7.74 L) = P₂(0.23 L) <----- Insert values
30.0312 = P₂(0.23 L) <----- Simplify left side
131 = P₂ <----- Divide both sides by 0.23
Answer:
0.297 °C
Step-by-step explanation:
The formula for the <em>freezing point depression </em>ΔT_f is
ΔT_f = iK_f·b
i is the van’t Hoff factor: the number of moles of particles you get from a solute.
For glucose,
glucose(s) ⟶ glucose(aq)
1 mole glucose ⟶ 1 mol particles i = 1
Data:
Mass of glucose = 10.20 g
Mass of water = 355 g
ΔT_f = 1.86 °C·kg·mol⁻¹
Calculations:
(a) <em>Moles of glucose
</em>
n = 10.20 g × (1 mol/180.16 g)
= 0.056 62 mol
(b) <em>Kilograms of water
</em>
m = 355 g × (1 kg/1000 g)
= 0.355 kg
(c) <em>Molal concentration
</em>
b = moles of solute/kilograms of solvent
= 0.056 62 mol/0.355 kg
= 0.1595 mol·kg⁻¹
(d) <em>Freezing point depression
</em>
ΔT_f = 1 × 1.86 × 0.1595
= 0.297 °C
An endothermic reaction is when the energy is absorbed, while an exothermic reaction releases energy.
I'd love to help, but you forgot to add the question.
