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densk [106]
3 years ago
5

What is the volume in liters of 321 g of a liquid with a density of 0.84 g/mL?

Chemistry
2 answers:
garri49 [273]3 years ago
7 0
If we have 321 grams of a liquid, and the density is 0.84 g/mL, then we can easily find the volume of the liquid. We just need to take this 0.84 and multiply that by the number of grams. If we do 321 * 0.84, we get 269.64 mL. This is the volume that this liquid has.Remember this equation for future problems: V = D*M. V meaning volume, D meaning density, and M meaning mass. I hope this helps.
Katena32 [7]3 years ago
5 0

<u>Answer:</u> The volume of liquid is 0.382 L

<u>Explanation:</u>

Density of a substance is defined as the ratio of its mass and volume. The equation used to determine density follows:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Density of liquid = 0.84 g/mL

Mass of liquid = 321 g

Putting values in above equation, we get:

0.84g/mL=\frac{321g}{\text{Volume of liquid}}\\\\\text{Volume of liquid}=\frac{321g}{0.84g/mL}=382.1mL

Converting this into liters, we use the conversion factor:

1 L = 1000 mL

So, 382.1mL\times \frac{1L}{1000mL}=0.382L

Hence, the volume of liquid is 0.382 L

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6 0
3 years ago
how do i solve these equations using proust's law? we haven't gone over balancing chem equations in class and our teacher doesn'
kow [346]

Answer:

Explanation:

Ok so Proust's Law, better known as the Law of Definite Proportions, states that the components of a compound always exist in a fixed ratio. This means that it does not matter what the source of the components are nor the coefficients in front of them. The ratio your teacher is referring to is most likely mass percent/ percent composition. This ratio is the amount of the component over the amount of the entire compound times 100%.

*I am not very familiar with this law so please do with my answers what you will :)*

7.)

First, you want to find the percent composition (aka mass composition) of Na in NaCl.

 45.89 g Na
---------------------  x 100%  =  39.3%
116.89 g NaCl

So, there must be 39.3% sodium in NaCl. You can find how much chlorine is in NaCl by subtracting that percent by 100 (to find the percent composition of chlorine) then multiplying it by the mass.

100% - 39.3% = 60.7%

60.7% / 100 = 0.0607

116.89 g NaCl x 0.0607 = 70.91 g Cl₂

You could also just subtract the mass of sodium from the mass of sodium chloride to find the mass of chlorine.

116.89 g NaCl - 45.89 g Na = 71 g Cl₂

8.)

10.57 g Mg + 6.96 g O₂ = 17.53 g MgO

  6.96 g O₂
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9.)

6.46 g Pb = 1 g O₂

68.54 g Pb  =  28.76 g O₂

68.54 g Pb / 28.76 g O₂ = 2.83 g ≠ 6.46 g

No, the two samples are not the same because the proportion of lead to oxygen is not the same for both samples. In the first sample, there is 6.46 g lead for every oxygen. In the second sample, there is 2.38 g lead for every oxygen.

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Answer:

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Explanation:

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Explanation:

The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.

This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated

The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.

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