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GalinKa [24]
2 years ago
7

Which of the following solutions would have [Fe3+] = 0.020 M?

Chemistry
1 answer:
Shalnov [3]2 years ago
7 0

Answer:

None of the given options

Explanation:

Let's go case by case:

A. No matter the volume, the concentration of Fe(NO₃)₃ (and thus of [Fe³⁺] as well) is 0.050 M.

B. We can calculate the moles of Fe₂(SO₄)₃:

  • 0.020 M * 0.80 L = 0.016 mol Fe₂(SO₄)₃

Given that there are two Fe⁺³ moles per Fe₂(SO₄)₃ mol, in the solution we have 0.032 moles of Fe⁺³. With that information in mind we <u>can calculate [Fe⁺³]</u>:

  • 0.032 mol Fe⁺³ / 0.80 L = 0.040 M

C. Analog to case A., the molar concentration of Fe⁺³ is 0.040 M.

D. Similar to cases A and C., [Fe⁺³] = 0.010 M.

Thus none of the given options would have [Fe⁺³] = 0.020 M.

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Calculate the molality of 75.0 grams of MgCl2 (molar mass=95.21 g/mol) dissolved in 500.0 g of solvent.
nordsb [41]

<u>Answer:</u> The molality of magnesium chloride is 1.58 m

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (magnesium chloride) = 75.0

M_{solute} = Molar mass of solute (magnesium chloride) = 95.21 g/mol  

W_{solvent} = Mass of solvent = 500.0 g

Putting values in above equation, we get:

\text{Molality of }MgCl_2=\frac{75.0\times 1000}{95.21\times 500.0}\\\\\text{Molality of }MgCl_2=1.58m

Hence, the molality of magnesium chloride is 1.58 m

5 0
3 years ago
Read 2 more answers
PLEASE HELP MEEE!!!!
Assoli18 [71]
It would be 35.8 Calories or calories. Not sure about that part. Hope this helps though.
3 0
3 years ago
If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p
Lapatulllka [165]

Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

as

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl  = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl  = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

3 0
3 years ago
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What is the percent composition for the compound NaBr? (8C)
ehidna [41]

Answer:

the answer is D

Explanation:

percentage composition=  mole of the substance divided by the total molar mass of the compound multiplied by 100.

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3 years ago
An atom bonds with another atom. What is the best classification for this reaction?
defon
Chemical would be the correct answer
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