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1 year ago

Based on the collision theory, which factors are likely to increase the rate of reaction? Select all that apply.

Chemistry

1 answer:

Alexus [3.1K]1 year ago

7 0

Answer:

Option a

Explanation:

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What is the mass of arsenic that is contained in a 50.0 g sample if the concentration is 5 ppm?

faltersainse [42]

Answer: The mass of arsenic found in the sample is 0.25 mg

Explanation:

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

Both the masses are in grams.

We are given:

Concentration of arsenic = 5 ppm

Mass of sample = 50 g

Putting values in above equation, we get:

$5=\frac{\text{Mass of arsenic}}{50g}\times 10^6\\\\\text{Mass of arsenic}=\frac{5\times 50}{10^6}=0.00025g=0.25mg$

Conversion factor used: 1 g = 1000 mg

Hence, the mass of arsenic found in the sample is 0.25 mg

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3 years ago

Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Ass

denis23 [38]

Answer:

607 ppm

Explanation:

In this case we can start with the ppm formula:

$ppm=\frac{mg~of~solute}{Litters~of~solution}$

If we have a solution of 0.0320 M, we can say that in 1 L we have 0.032 mol of $F^-$ , because the molarity formula is:

$M=\frac{mol}{L}$

In other words:

$0.0320~M=\frac{mol}{1~L}$

$mol=0.032~M*1~L=0.032~mol$

$1~L~of~Solution=0.0320~mol~of~solute$

If we use the atomic mass of $F$ (19 g/mol) we can convert from mol to g:

$0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g$

Now we can convert from g to mg (1 g= 1000 mg), so:

$0.607~g\frac{1000~mg}{1~g}=607~mg$

Finally we can divide by 1 L to find the ppm:

$ppm=\frac{607~mg}{1~L}=~607~ppm$

We will have a concentration of 607 ppm.

I hope it helps!

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3 years ago

How many electrons do carbon and oxygen share? I-0-I :0: Intro Done

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3 years ago

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A helium balloon at 23 C has a volume of 755 mL. The balloon is placed into a freezer kept at -10 C. Assuming pressure is consta

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Explanation:

T1/V1 = T2/V2

V2 = T2 x V1/T1

V2 = 296.15x535/319.15

V2 = 496.44mL

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