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Lady bird [3.3K]
3 years ago
9

Tia has a sample of pure gold (Au). She weighed the sample and the result was 62.4 grams. Tia wants to determine the number of a

toms in the sample. Calculate the number of atoms in 62.4 g of pure gold.
Chemistry
1 answer:
Marta_Voda [28]3 years ago
7 0
I hope you are able to find your answer through the guidance of this site as I have yet to cover this topic in chemistry: http://scientifictutor.org/1021/chem-how-to-convert-between-grams-and-molecules/
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a beer's law plot was generated and the trendline had a slope of 137.2 l/mol. the absorbance of the unknown phosphate sample was
Lorico [155]

The phosphate concentration of the unknown sample is <u>0.002 M</u>

<u />

Beer's law, the quantity of power absorbed or transmitted by using a solution is proportional to the solution's molar absorptivity and the awareness of the solute. In easy terms, a greater focused answer absorbs greater light than a more dilute solution does.

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1 year ago
1- A sample of gas is in a 3.00 L contain at a pressure of 740.0 mmHg. What is the new pressure of the sample if the container's
inna [77]

Considering the Boyle's law, the new pressure of the sample is 1,776 mmHg.

<h3>What is Boyle's law</h3>

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant.

Boyle's law states that the volume occupied by a given mass of gas at constant temperature is inversely proportional to the pressure. This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

P×V=k

Now it is possible to assume that you have a certain volume of gas V1 which is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and the following will be true:

P1×V1=P2×V2

<h3>New pressure</h3>

In this case, you know:

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Replacing in Boyle's law:

740 mmHg× 3 L=P2× 1.25 L

Solving:

P2= (740 mmHg× 3 L) ÷ 1.25 L

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