07.05.2020

what is the molar i ty of 4.1 liters of solution containing 33.8 grams of nacl? na:23 cl:35

. 4

Faq

Chemistry
Step-by-step answer
P Answered by PhD

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

\frac{2.0 L}{1}\frac{1.5 mol}{L}\frac{85.01 g}{1 mol}

= 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)

Chemistry
Step-by-step answer
P Answered by PhD

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

\frac{2.0 L}{1}\frac{1.5 mol}{L}\frac{85.01 g}{1 mol}

= 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)

Chemistry
Step-by-step answer
P Answered by PhD

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

Chemistry
Step-by-step answer
P Answered by PhD

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

Chemistry
Step-by-step answer
P Answered by PhD

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)

Chemistry
Step-by-step answer
P Answered by PhD

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)

Chemistry
Step-by-step answer
P Answered by Specialist

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:

\frac{203,9 moles+ Xmoles}{87,8L + Yliters}(1)

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L×\frac{1,498 kgsolution}{L}×\frac{60 kg H_{2}SO_{4}}{100kgSolution}×\frac{1kmol}{98,08kg H_{2}SO_{4}} = 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!

Chemistry
Step-by-step answer
P Answered by Specialist

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:

\frac{203,9 moles+ Xmoles}{87,8L + Yliters}(1)

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L×\frac{1,498 kgsolution}{L}×\frac{60 kg H_{2}SO_{4}}{100kgSolution}×\frac{1kmol}{98,08kg H_{2}SO_{4}} = 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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