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Zolol [24]
2 years ago
5

What is the density of an aqueous solution that is 50.0% KOH by mass and has a KOH concentration of 13.39 M?

Chemistry
1 answer:
Contact [7]2 years ago
7 0

Answer:

Density = 1.56 g/mL

Explanation:

We have two sort of concentrations in here, so let's combine them.

% by mass means grams of solute in 100 g of solution

M means, moles of solute in 1L of solution.

We know that our solute is KOH (Molar mass = 56.1 g/mol)

We convert the moles of solute to grams:

56.1 g/mol . 13.9 moles = 779.79 g

Now, we know that 50 g of solute are in 100 g of solution

Then, 779.79 g of solute will be contained in (100 . 779.79)/50 = 1559.58 g of solution

Density = mass / volume

1559.58 g of solution / 1000 mL of solution = 1.56 g/mL

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At Jim's auto shop, it takes him minutes to do an oil change and minutes to do a tire change. Let be the number of oil changes h
Juli2301 [7.4K]

Answer:

The total time that Jim needs to change x oil changes and y tire changes is less than 180 min.

The time needed for x oil changes is 12 * x.

The time needed for y tire changes is 18 * y.

The total time is the sum of the above times and needs to be less than 180 that is

12 * x + 18 * y < 180 divide both sides of equation by 6

12/6 * x + 18/6*y < 180/6

2*x + 3*y < 30

2*x < 30 - 3*y divide both sides by 2 to get the inequality for x

x < 30/2 - 3/2*y = 15 - 1.5 y < 15 that is x < = 15

2*x + 3*y < 30

3*y < 30 - 2*x divide both sides by 3 to get the inequality for y

y < 30/3 - 2/3 *x = 10 - 2/3*x < 10 that is y < = 10

Also we can write x + y < x+ 3/2 * y < 15.

Explanation:

Jim's can do not more then 5 oil changes and not more then 10 tire changes or all together she can do not more then 15 total of oil and tire changes.

5 0
3 years ago
What do you get when you mix lemons with gun powder
Viefleur [7K]
A very disgusting type of lemonade 

6 0
3 years ago
decomposition of hydrogen peroxide produce water and oxygen . calculate the volume of O2 formed from the decomposition of 150 mL
Ugo [173]

Explanation:

2H2O2 => 2H2O + O2

Moles of hydrogen peroxide = 0.150dm³ * (0.02mol/dm³) = 0.003mol .

Moles of oxygen = 0.0015mol.

Volume of oxygen = 0.0015mol * (22.4dm³/mol) = 0.0336dm³.

5 0
2 years ago
Which of the following could be said about ecological footprints?
Thepotemich [5.8K]
<span>A. Exact ecological footprints are often difficult to calculate, but estimates can be useful in comparing populations.

</span>Which of the following could be said about ecological footprints? <u /> <u>Exact ecological footprints are often difficult to calculate, but estimates can be useful in comparing populations.</u><u />


NOT:
b. Ecological footprints can't be used to determine carrying capacity.
C. Ecological footprints don't take into account resources needed to absorb and manage wastes.
<span>D. The average ecological footprints for various countries are nearly identical.</span>
6 0
3 years ago
A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot
kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

p_A=\chi_A\times P_T

p_{O_2} = partial pressure of O_2 = ?

\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}

P_{T} = total pressure of mixture  = 525 mmHg

{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles

{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles

Total moles = 1.94 + 1.35 = 3.29 moles

\chi_{O_2}=\frac{1.94}{3.29}=0.59

p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0
3 years ago
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