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nevsk [136]
3 years ago
11

8.5ml of a sample of sea water solution was added to a 44.317g evaporating dish the combination weighted 52.987g after evaporati

ng the dish and content weighed 44.599g
Chemistry
1 answer:
Ratling [72]3 years ago
5 0

Answer:  There is no question, but we can calculate a couple of items:

Density of sea water sample = (52.987g-44.317g)/8.5ml

Inorganic content of sample (mostly salts) = (44.599g-44.317g)/(52.987g-44.317g) x 100% = percent inorganics in water sample

Explanation:

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Question 1 :

V1/T1 = V2/T2

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Question 2 :

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Question 3 :

Given that:

Volume of gas V = 4.80L

(since 1 liter = 1dm3

4.80L = 4.80dm3)

Temperature T = 62°C

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Apply ideal gas equation

pV = nRT

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Question 4 :

Volume of gas V = 3.47L

(since 1 liter = 1dm3

3.47L = 3.47dm3)

Temperature T = 85.0°C

Convert Celsius to Kelvin

(85.0°C + 273 = 358K)

Pressure P = ?

Number of moles of gas N = 0.100 mole

Apply ideal gas equation

pV = nRT

p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)

p x 3.47dm3 = 0.29 atm dm3

p = (0.29 atm dm3 / 3.47 dm3)

p = 0.085 atm

If 1 atm = 760 mm Hg

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