No. of moles of calcium chloride = molarity × volume of solution in L
No. of moles of calcium chloride = 2.5 ×0.5 = 1.25 mole
No. of moles of calcium chloride = mass of calcium chloride / molar mass of calcium chloride
1.25 mole = mass of calcium chloride / 110.98 g/mol
mass of calcium chloride = 1.25 ×110.98 = 139 g
Answer:
The initial temperature of the metal is 84.149 °C.
Explanation:
The heat lost by the metal will be equivalent to the heat gain by the water.
- (msΔT)metal = (msΔT)water
-32.5 grams × 0.365 J/g°C × ΔT = 105.3 grams × 4.18 J/g °C × (17.3 -15.4)°C
-ΔT = 836.29/12.51 °C
-ΔT = 66.89 °C
-(T final - T initial) = 66.89 °C
T initial = 66.89 °C + T final
T initial = 66.89 °C + 17.3 °C
T initial = 84.149 °C.
Ba 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² → [Xe]6s²
Ba - 2e⁻ → Ba⁺² [Xe]
<u>Given:</u>
% Al = 35.94
% S = 64.06
<u>To determine:</u>
Empirical formula of a compound with the above composition
<u>Explanation:</u>
Atomic wt of Al = 27 g/mol
Atomic wt of S = 32 g/mol
Based on the given data, for 100 g of the compound: Mass of Al = 35.94 g and mass of S = 64.06 g
# moles of Al = 35.94/27 = 1.331
# moles of S = 64.06/32 = 2.002
Divide by the smallest # moles:
Al = 1.331/1.331 = 1
S = 2.002/1,331 = 1.5 ≅ 2
Empirical formula = AlS₂
Answer: The partial pressure of oxygen in the mixture is 321 mm Hg
Explanation:
According to Dalton's law, the total pressure is the sum of individual pressures.
Given : 
= total pressure of gases = 752 mm Hg
= partial pressure of Helium = 234 mm Hg
= partial pressure of nitrogen = 197 mm Hg
= partial pressure of oxygen = ?
Putting in the values we get:
The partial pressure of oxygen in the mixture is 321 mm Hg
