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solmaris [256]
3 years ago
12

A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan

ge when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Chemistry
1 answer:
Dimas [21]3 years ago
7 0

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = \frac{18.0 g}{60 g/mol}=0.3 mol

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}

Molarity of the urea solution ;

M=\frac{0.3 mol}{0.200 L}=1.50 M

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

m=d\times V=0.95 g/mL\times 200.0 mL=190 g

m = 190 g = 190 \times 0.001 kg = 0.19 kg

(1 g = 0.001 kg)

Molality of the urea solution ;

Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}

m=\frac{0.3 mol}{0.19 kg}=1.58 m

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

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Octane has a density of 0.703 g/ml. Calculate the mass of CO2(g) produced by burning one
STatiana [176]

The mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

The given parameters:

  • <em>Density of the octane, ρ = 0.703 g/ml</em>
  • <em>Volume of octane, v = 3.79 liters</em>

<em />

The mass of the octane burnt is calculated as follows;

m = \rho V\\\\m = 0.703 \ \frac{g}{ml} \times 3.79 \ L \ \frac{1000 \ ml}{L} \\\\m = 2,664.37 \ g

The combustion reaction of octane is given as;

2C_8H_{18} +  \ 25O_2 \ --> \ 16CO_2 \ + \ 18H_2O

From the reaction above:

228.46 g of octane -------------------> 704 g of  CO₂ gas

2,664.37 of octane --------------------> ? of CO₂ gas

= \frac{2,664.37 \times 704}{228.46} \\\\= 8,210.3 \ g\\\\= 8.21 \ kg

Thus, the mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

Learn more about combustion of organic compounds here: brainly.com/question/13272422

8 0
2 years ago
HELP PLSSSS
den301095 [7]

Answer:

Groups like the Halogens, which include Chlorine and Flourine, share similar properties both behaviorally and structurally. The Periodic Table is essentially a bunch of patterns and trends and the groups (like the one with Sodium and Potassium) were grouped together because of these similarities.

Explanation:

https://en.wikipedia.org/wiki/Alkali_metal

8 0
2 years ago
Compounds X and Y are both C6H13Cl compounds formed in the radical chlorination of 3-methylpentane. Both X and Y undergo base-pr
zubka84 [21]

Answer:

Y is a 3-chloro-3-methylpentane.

The structure is shown in the figure attached.

Explanation:

The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).

The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.

6 0
3 years ago
(e) Another student investigated the rate of a different reaction
vodka [1.7K]

Answer:

0.07 g/s.

Explanation:

From the question given above, the following data were obtained:

Mass lost = 9.85 g

Time taken = 2 min 30 s

Mean rate =?

Next, we shall convert 2 min 30 s to seconds (s). This can be obtained as follow:

1 min = 60 s

Thus,

2 min = 2 × 60 = 120 s

Therefore,

2 min 30 s = 120 s + 30 s = 150 s

Finally, we shall determine the mean rate of the reaction. This can be obtained as illustrated below:

Mass lost = 9.85 g

Time taken = 150 s

Mean rate =?

Mean rate = mass lost / time taken

Mean rate = 9.85 / 150

Mean rate = 0.07 g/s

Therefore, the mean rate of the reaction is 0.07 g/s

4 0
3 years ago
How many milliseconds (ms) are there in 3.5 seconds (s)?
ivolga24 [154]

Answer:

13 hundred

Explanation:

4 0
3 years ago
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