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lora16 [44]
3 years ago
10

Help with chemistry, please

Chemistry
2 answers:
Hitman42 [59]3 years ago
3 0
The cubes have only the same volume, so the answer is c.
maks197457 [2]3 years ago
3 0
The answer is the same volume
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In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-bu
Novosadov [1.4K]

Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles

We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol =  0.810 g/mL

Volume of  3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol =  mass/molar mass =  7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004

7 0
2 years ago
What happens when an electron moves from n=2 to n=5?
Aleksandr-060686 [28]

Answer:

Explanation:

10?

4 0
2 years ago
Ricardo is working on a refrigeration system and needs to know the fluid pressure of a water tank. He knows that the tank is 12
hjlf

Answer:

I also have this question but see the 69 page of ur question book

8 0
2 years ago
30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr
Alenkasestr [34]

Answer: Thus concentration of Ca(OH)_2 in mol/dm^3  is 0.011 and in g/dm^3 is 0.814

Explanation:

To calculate the concentration of Ca(OH)_2, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3         1cm^3=0.001dm^3

Putting values in above equation, we get:

1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3

The concentration in g/dm^3 is 0.011mol/dm^3\times 74g/mol=0.814g/dm^3

Thus concentration of Ca(OH)_2 is 0.011mol/dm^3 and 0.814g/dm^3

4 0
3 years ago
I need help with this
lubasha [3.4K]
The answer is untestable
5 0
3 years ago
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