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Keith_Richards [23]
3 years ago
13

What is the point of previewing? a. to get the gist of the text you don’t have read the whole text c. to gather as much informat

ion on the text as possible so you can better understand the text when read. b. to save time d. none of the above Please select the best answer from the choices provided A B C D
Chemistry
1 answer:
Murljashka [212]3 years ago
7 0

C. TO GATHER AS MUCH INFORMATION ON THE TEXT AS POSSIBLE SO YOU CAN BETTER UNDERSTAND THE TEXT WHEN YOU READ


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Calculate the mass percent of Cl in Freon -112 (C2 Cl4 F2), a CFC refrigerant
Karolina [17]

Answer:

The answer to your question is:  69.6 %

Explanation:

Freon -112 (C₂Cl₄F₂)

MW = (12 x 2) + (35.5 x 4) + (19 x 2)

      = 24 + 142 + 38

      = 204 g

                       204 g of C₂Cl₄F₂  -----------------  100%

                       142 g                     -----------------   x

                      x = (142 x 100 ) / 204

                      x = 69.6 %

5 0
3 years ago
Read 2 more answers
ASAP
nevsk [136]

I’m pretty sure it’s the second answer B

5 0
3 years ago
If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount
nirvana33 [79]

<u>Answer:</u> The volume of water required is 398 mL

<u>Explanation:</u>

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

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of solute (manganese (II) nitrate tetrahydrate) = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Molarity of solution = 0.16 M

Putting values in above equation, we get:

0.16M=\frac{16g\times 1000}{251g/mol\times \text{Volume of solution}}\\\\\text{Volume of solution}=398mL

Hence, the volume of water required is 398 mL

7 0
3 years ago
how many grams of phosphorus (P4) react with 35.5L of O2 at STP to form solid tetraphosphorus decaoxide
algol [13]

Answer:

mass P4 = 35.998 g

Explanation:

  • P4 + 5O2 → P4O10

∴ STP: P = 1 atm; T = 298 K

∴ V O2= 35.5 L

⇒ nO2 = P.V / R.T

∴ R = 0.082 atm.L/K.mol

⇒ nO2 = ((1 atm)×(35.5L))/((0.082 atm.L/K.mol)(298K))

⇒ nO2 = 1.453 mol O2

⇒ mol P4 = (1.453 molO2)×(mol P4/ 5molO2) = 0.2906 mol P4

∴ Mw P4 = 123.895 g/mol

⇒ mass P4 = (0.2906 mol P4)×(123.895 g/mol) = 35.998 g P4

4 0
3 years ago
Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to th
Luba_88 [7]

Answer:

The percent yield of the reaction is 82%

Explanation:

First step: make the chemist equation.

2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)

As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3

Second step: Find out the moles in the reactant.

Molar weight Fe2O3: 159.7 g/m

Mass / Molar weight = moles

50 g /159.7 g/m = 0.313 moles

Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.

1 mol Fe2O3 ____ 2Fe

0.313 mol Fe2O3 ____ 0.626 moles

Molar weight Fe = 55.85 g/m

Moles . molar weight = mass

55.85g/m . 0.626m = 34.9 grams

This will be the 100% yield of the reaction but we only made 28.65 g

34.9 g ____ 100%

28.65 g ____ 82.09 %

3 0
3 years ago
Read 2 more answers
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