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Ierofanga [76]
3 years ago
13

How many grams of a 14.0% (w/w) sugar solution contain 62.5 of sugar?

Chemistry
1 answer:
Delicious77 [7]3 years ago
3 0

Answer:

m_{solution}=446.4g

Explanation:

Hello,

In this case, by-mass percent is computed in terms of the mass of the solute and the mass of the solvent as shown below:

\% m=\frac{m_{solute}}{m_{solution}} *100\%

Thus, solving for the mass of the solution, we obtain:

\frac{x}{y} m_{solution}=\frac{m_{solute}}{\% m}*100\%\\ \\m_{solution}=\frac{62.5g}{14.0\%}*100\% \\\\m_{solution}=446.4g

Regards.

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kirza4 [7]
The Correct Answer Is 3.2
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Hydrogen peroxide decomposes to form water and oxygen gas
EleoNora [17]
True? If that’s what you are asking. It does decompose to that.
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An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the
Misha Larkins [42]

Answer:

The correct answer is 199.66 grams per mole.

Explanation:

Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,  

R1/R2 = √ M2/√ M1

Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.  

Rate Q/Rate N2 = √M of N2/ √M of Q

The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67

Now putting the values we get,  

rate of N2/2.67/rate of N2 = √28/ √M of Q

√M of Q = √ 28 × 2.67

M of Q = (√ 28 × 2.67)²

M of Q = 199.66 grams per mole

3 0
3 years ago
Question 7 of 10
nirvana33 [79]

Answer:

I think the answer is D. Temperature affects only the rate of reaction.

7 0
3 years ago
A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure
Alex

Answer:

B.3/5p

Explanation:

For this question, we have to remember <u>"Dalton's Law of Partial Pressures"</u>. This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a <em>proportional relationship between moles and pressure</em>. In other words, more moles indicate more pressure and vice-versa.

P_i=P_t_o_t_a_l*X_i

Where:

P_i=Partial pressure

P_t_o_t_a_l=Total pressure

X_i=mole fraction

With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:

<u>moles of hydrogen gas</u>

The molar mass of hydrogen gas (H_2) is 2 g/mol, so:

6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2

<u>moles of oxygen gas</u>

The molar mass of oxygen gas (O_2) is 32 g/mol, so:

64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2

Now, total moles are:

Total moles = 2 + 3 = 5

With this value, we can write the partial pressure expression for each gas:

P_H_2=\frac{3}{5}*P_t_o_t_a_l

P_O_2=\frac{2}{5}*P_t_o_t_a_l

So, the answer would be <u>3/5P</u>.

I hope it helps!

5 0
3 years ago
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