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Liula [17]
3 years ago
5

Chem lab report is attached to the question plz help.

Chemistry
2 answers:
DIA [1.3K]3 years ago
3 0

Answer: When I opened the doc it didn't work..

Explanation:

Mariulka [41]3 years ago
3 0

Answer:

I can’t see your problem it doesn’t work

Explanation:u should download doc

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15.0 g of Fe and 25.0 g of sand are added to 250.0 g of water. a. Determine the percent mass of Fe, sand, and water in the mixtu
lana [24]

Answer:

A. percentage mass of iron = 5.17%

percentage mass of sand  = 8.62%

percentage mass of water = 86.205%

B. (Iron + sand + water) -------> ( iron + sand) ------> sand

C. The step of separation of iron and sand

Explanation:

A. Percentage mass of the mixtures:

Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g

percentage mass of iron = 15/290 * 100% = 5.17%

percentage mass of sand = 25/290 * 100% = 8.62%

percentage mass of water = 250/290 * 100% = 86.205%

B. Flow chart of separation procedure

(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand

C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.

5 0
2 years ago
An atom was observed with +3 charge, n = 38, and e = 28. What is the mass number of the atom?
Finger [1]

Answer: 66

Explanation: To find the mass number you would have to add the number of protons and the number of neutrons.

3 0
2 years ago
if 30.0L of oxygen are cooled from 200 degrees celsius to 1 degree celsius at constant pressure, what is the new volume of oxyge
Sergio039 [100]

Answer: 18.65L

Explanation:

Given that,

Original volume of oxygen (V1) = 30.0L

Original temperature of oxygen (T1) = 200°C

[Convert temperature in Celsius to Kelvin by adding 273.

So, (200°C + 273 = 473K)]

New volume of oxygen V2 = ?

New temperature of oxygen T2 = 1°C

(1°C + 273 = 274K)

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

30.0L/473K = V2/294K

To get the value of V2, cross multiply

30.0L x 294K = 473K x V2

8820L•K = 473K•V2

Divide both sides by 473K

8820L•K / 473K = 473K•V2/473K

18.65L = V2

Thus, the new volume of oxygen is 18.65 liters.

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3 years ago
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What part is the independent variable and what part is the dependent variable in the scenario: The blood pressure of a soldier i
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