The following questions are answered below
Explanation:
1.a. 1.0 liter of a 1.0 M mercury (II) chloride (HgCl2) solution
1.0 L 1 mol 271.49 g
L 1 L 1 mol
= 271.49 g
= HgCl₂ = 271.49 g HgCl₂
1.b. 2.0 liters of a 1.5 M sodium nitrate (NaNO3) solution
= 255.03 g
= 255 g NaNo₃
1.c. 5.0 liters of a 0.1 M Ca(OH)2 solution
= 37 g Ca(OH)2
1.d. 100 mL of a 0.5 M (NH4)3PO4 solution
= 7.5 g (NH₄)₃PO₄
2. To find the molarity of the following solutions
2.a. 12 g of lithium hydroxide (LiOH) in 1.0 L of solution
= 0.50 m (LiOH)
2.b. 198 g of barium bromide (BaBr2) in 2.0 L of solution
= 0.33 m (BaBr₂)
2.c. 54 g of calcium sulfide (CaS) in 3.0 L of solution
= 0.25 m (CaS)
3. To find the volume of each solution
3.a. 1.0 M solution containing 85 g of silver nitrate (AgNO3)
= 0.50 L (AgNO₃)
3.b. 0.5 M solution containing 250 g of manganese (II) chloride (MnCl2)
= 4.0 L MnCl2
3.c. 0.4 M solution containing 290 g of aluminum nitrate (Al(NO3)3)
= 3.4 L (Al(NO₃)₃)
The following questions are answered below
Explanation:
1.a. 1.0 liter of a 1.0 M mercury (II) chloride (HgCl2) solution
1.0 L 1 mol 271.49 g
L 1 L 1 mol
= 271.49 g
= HgCl₂ = 271.49 g HgCl₂
1.b. 2.0 liters of a 1.5 M sodium nitrate (NaNO3) solution
= 255.03 g
= 255 g NaNo₃
1.c. 5.0 liters of a 0.1 M Ca(OH)2 solution
= 37 g Ca(OH)2
1.d. 100 mL of a 0.5 M (NH4)3PO4 solution
= 7.5 g (NH₄)₃PO₄
2. To find the molarity of the following solutions
2.a. 12 g of lithium hydroxide (LiOH) in 1.0 L of solution
= 0.50 m (LiOH)
2.b. 198 g of barium bromide (BaBr2) in 2.0 L of solution
= 0.33 m (BaBr₂)
2.c. 54 g of calcium sulfide (CaS) in 3.0 L of solution
= 0.25 m (CaS)
3. To find the volume of each solution
3.a. 1.0 M solution containing 85 g of silver nitrate (AgNO3)
= 0.50 L (AgNO₃)
3.b. 0.5 M solution containing 250 g of manganese (II) chloride (MnCl2)
= 4.0 L MnCl2
3.c. 0.4 M solution containing 290 g of aluminum nitrate (Al(NO3)3)
= 3.4 L (Al(NO₃)₃)
2 — 1.6 mol·L⁻¹
Explanation:
1. Write the chemical equation for the reaction.
Ca(OH)₂ + 2HCl ⟶ CaCl₂ + 2H₂O
2. Calculate the mass of Ca(OH)₂.
V = 100 mL
1000 mL of solution contain 30 g Ca(OH)₂ Calculate the mass of Ca(OH)₂
Mass of Ca(OH)₂ = 100 × 30/1000
Mass of Ca(OH)₂ = 3.00 g Ca(OH)₂
3. Calculate the moles of Ca(OH)₂
1 mol Ca(OH)₂ = 74.09 g Calculate the moles of Ca(OH)₂
Moles of Ca(OH)₂ = 3.00 × 1/74.09
Moles of Ca(OH)₂ = 0.0405 mol Ca(OH)₂
4. Calculate the moles of HCl
2 mol HCl ≡ 1 mol Ca(OH)₂ Calculate the moles of HCl
Moles of HCl = 0.0405 × 2/1
Moles of HCl = 0.0810 mol HCl
5. Calculate the molar concentration of the HCl
c = n/V
V = 50 mL = 0.050 L Calculate the concentration
c = 0.0810/0.050
c = 1.6 mol·L⁻¹
2 — 1.6 mol·L⁻¹
Explanation:
1. Write the chemical equation for the reaction.
Ca(OH)₂ + 2HCl ⟶ CaCl₂ + 2H₂O
2. Calculate the mass of Ca(OH)₂.
V = 100 mL
1000 mL of solution contain 30 g Ca(OH)₂ Calculate the mass of Ca(OH)₂
Mass of Ca(OH)₂ = 100 × 30/1000
Mass of Ca(OH)₂ = 3.00 g Ca(OH)₂
3. Calculate the moles of Ca(OH)₂
1 mol Ca(OH)₂ = 74.09 g Calculate the moles of Ca(OH)₂
Moles of Ca(OH)₂ = 3.00 × 1/74.09
Moles of Ca(OH)₂ = 0.0405 mol Ca(OH)₂
4. Calculate the moles of HCl
2 mol HCl ≡ 1 mol Ca(OH)₂ Calculate the moles of HCl
Moles of HCl = 0.0405 × 2/1
Moles of HCl = 0.0810 mol HCl
5. Calculate the molar concentration of the HCl
c = n/V
V = 50 mL = 0.050 L Calculate the concentration
c = 0.0810/0.050
c = 1.6 mol·L⁻¹
See explanations
Explanation:
a.Molarity = moles/Volume in Liters = 5moles/2Liters = 2.5M in NaCl
b.Freezing Pt Depression
1.Sprinkling salt on icy surfaces
2.Using antifreeze in automobile cooling systems
3.Not an application
4.Using salt to make ice cream
c.pOH = -log[OHˉ] = -log(1x10ˉ¹⁰) = -(-10) = 10 => pH = 14 – pOH = 14 – 10 = 4
d.H₂O + NH₃ => NH₄⁺ + OHˉ => Bronsted Acid is H₂O (proton donor)
See explanations
Explanation:
a.Molarity = moles/Volume in Liters = 5moles/2Liters = 2.5M in NaCl
b.Freezing Pt Depression
1.Sprinkling salt on icy surfaces
2.Using antifreeze in automobile cooling systems
3.Not an application
4.Using salt to make ice cream
c.pOH = -log[OHˉ] = -log(1x10ˉ¹⁰) = -(-10) = 10 => pH = 14 – pOH = 14 – 10 = 4
d.H₂O + NH₃ => NH₄⁺ + OHˉ => Bronsted Acid is H₂O (proton donor)
The feed ratio (liters 20% solution/liter 60% solution) is 3,08
Explanation:
In this problem you have a 20,0 wt% H₂SO₄ and a 60,0 wt% H₂SO₄ solutions.
100 kg of 20% solution are 100kg/1,139 kg/L = 87,8 L
100kg×20wt% = 20 kg H₂SO₄. In moles:
20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ 203,9 mol
The final molarity 4,00M comes from:
(1)
Where X moles and Y liters comes from 60,0 wt% H₂SO₄
100 L of 60,0 wt% H₂SO₄ are:
100L××× = 0,9164 kmolH₂SO₄ ≡ 916,4 moles
That means:
X/Y = 916,4/100 = 9,164 (2)
Replacing (2) in (1):
Y(liters of 60,0 wt% H₂SO₄) = 28,52 L
Thus, feed ratio (liters 20% solution/liter 60% solution):
87,8L/28,52L = 3,08
I hope it helps!
The feed ratio (liters 20% solution/liter 60% solution) is 3,08
Explanation:
In this problem you have a 20,0 wt% H₂SO₄ and a 60,0 wt% H₂SO₄ solutions.
100 kg of 20% solution are 100kg/1,139 kg/L = 87,8 L
100kg×20wt% = 20 kg H₂SO₄. In moles:
20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ 203,9 mol
The final molarity 4,00M comes from:
(1)
Where X moles and Y liters comes from 60,0 wt% H₂SO₄
100 L of 60,0 wt% H₂SO₄ are:
100L××× = 0,9164 kmolH₂SO₄ ≡ 916,4 moles
That means:
X/Y = 916,4/100 = 9,164 (2)
Replacing (2) in (1):
Y(liters of 60,0 wt% H₂SO₄) = 28,52 L
Thus, feed ratio (liters 20% solution/liter 60% solution):
87,8L/28,52L = 3,08
I hope it helps!
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