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Deffense [45]
3 years ago
14

How many grams of glucose are needed to prepare 400. ml of a 2.0%(m/v) glucose solution?

Chemistry
2 answers:
tatuchka [14]3 years ago
8 0
8 grams

A <span>2.0%m/v</span> (mass/volume) solution is 2 grams of solute in 100 mL of final solution.

So for 400 mL of solution, you require

<span>400mL⋅<span>2 g<span>100mL</span></span>=<span>8 g</span></span>

erma4kov [3.2K]3 years ago
4 0

Answer:8 grams of glucose are needed to prepare 400. ml of a 2.0%(m/v) glucose solution.

Explanation:

Volume of the solution = 400 mL

Mass by volume percentage of the solution = 2%

Mass of the glucose = m

The mass by volume percent is given by formula ;

(m/v\%)=\frac{\text{mass of the solute}}{\text{Volume of the solution}}\times 100

2\%=\frac{m}{400}\times 100

m = 8 g

8 grams of glucose are needed to prepare 400. ml of a 2.0%(m/v) glucose solution.

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7. How many chlorine atoms are in a 34.2 g sample of dichlorine pentoxide?
Reptile [31]

Answer:

The answer to your question is 1.36 x 10²³ atoms

Explanation:

Data

number of atoms = ?

mass of the sample = 34.2 g

Molecule = Cl₂O₅

Process

1.- Calculate the molar mass of Cl₂O₅

Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g

2.- Calculate the atoms of Cl₂O₅

                     151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms

                       34.2 g of Cl₂O₅ ------------ x

                          x = (34.2 x 6.023 x 10²³) / 151

                          x = 1.36 x 10²³ atoms

4 0
3 years ago
How many chlorine atoms are in each set?<br> six calcium chloride formula units.
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Answer:

tae ko malutong

Explanation:

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3 0
3 years ago
An ideal gas has a volume of 5.00L under a pressure of 1.00 atm and a temperature of 10.0 K. If the temperature is changed to 10
tigry1 [53]

Answer:

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Explanation:

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7 0
3 years ago
How many bromine atoms are present in 37.9 g of CH2Br2?
VARVARA [1.3K]

Answer:

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

                   174 g of CH2Br2 ------------------  160 g of Br2

                   37.9 g of CH2Br2   ---------------     x

                x = 37.9 x 160/174 = 34.85 g of Br

                      1 mol of Br -----------------   160 g Br2

                         x              ----------------    174 g Be2

               x = 174 x 1 /160 = 1.088 mol of Br2

                1 mol of Br -----------------  6.023 x 10 ²³ atoms

            1.088 mol of Br -------------    x

                    x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms

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3 years ago
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