1) Chemical reaction:
2(CH₃COO)₃Fe + 3MgCrO₄ → Fe₂(CrO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgCrO₄) = 10,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol.
n(MgCrO₄) = m(MgCrO₄) ÷ M(MgCrO₄).
n(MgCrO₄) = 10 g ÷ 140,3 g/mol.
n(MgCrO₄) = 0,071 mol; limiting reagens.
From chemical reaction: n(MgCrO₄) : n((CH₃COO)₂Mg) = 3 : 3.
n((CH₃COO)₂Mg) = 0,071 mol.
m((CH₃COO)₂Mg) = 0,071 mol · 142,4 g/mol.
n((CH₃COO)₂Mg) = 10,11 g.
2) Chemical reaction:
2(CH₃COO)₃Fe + 3MgSO₄ → Fe₂(SO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgSO₄) = 15,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol; limiting ragens.
n(MgSO₄) = m(MgSO₄) ÷ M(MgSO₄).
n(MgSO₄) = 15 g ÷ 120,36 g/mol.
n(MgSO₄) = 0,125 mol; limiting reagens.
From chemical reaction: n(CH₃COO)₃Fe) : n((CH₃COO)₂Mg) = 2 : 3.
n((CH₃COO)₂Mg) = 0,064 mol · 3/2.
n((CH₃COO)₂Mg) = 0,096 mol.
m((CH₃COO)₂Mg) = 0,096 mol · 142,4 g/mol.
m((CH₃COO)₂Mg) = 13,7 g.
Answer:
Explanation:
Chemical reactions involve combining different substances. The chemical reaction produces a new substance with new and different physical and chemical properties. Matter is never destroyed or created in chemical reactions. The particles of one substance are rearranged to form a new substance.
7.8 x 10^-3 = 0.0078 lb/qt
1lb = 454 g
0.0078 / 454 = 1.718061 x 10^-5 g/qt
Answer:
Explanation:
4 = Given data:
Initial volume = 400 mL
Initial pressure = 450 torr
Initial temperature = 210 K
Final temperature = ?
Final volume = 1500 mL
Final pressure = 800 torr
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁/ P₁V₁
T₂ = 800 torr × 1500 mL × 210 K / 450 torr × 400 mL
T₂ = 252000,000 K / 180000
T₂ = 1400 K
5 = Given data:
Initial volume of gas = 4.5 L
Initial temperature = 25°C (25 + 273 = 298 k)
Final temperature = 25°C×3= 75°C (75+273 = 348 k)
Final volume = ?
Solution:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 4.5 L × 348 K / 298 k
V₂ = 1566 L.K / 298 K
V₂ = 5.3 L
Extra credit:
Given data:
Initial volume = 356 cm³ or 356 mL
Initial pressure = 105000 pa
Initial temperature = 23 °C
Final temperature = ?
Final volume = 560 dL
Final pressure = 36 psi
Formula:
Final volume = 560×100 = 5600 mL
Initial temperature = 23 °C ( 273 + 23 = 296 K)
Final pressure = 36 × 6895 = 248220 Pa
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁/ P₁V₁
T₂ = 248220 Pa × 5600 mL × 296 K /105000 pa × 356 mL
T₂ = 411449472,000 K / 37380000
T₂ = 11007.21 K