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Fudgin [204]
3 years ago
14

CHEM HELP!

Chemistry
1 answer:
sweet [91]3 years ago
4 0

So let's convert this amount of mL to grams:

\frac{13.6g}{1mL}*1.2mL=16.32g

Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):

\frac{1mole}{200.59g}*16.32g=8.135*10^{-2}mol

Then we need to convert moles to atoms using Avogadro's number:

\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms

So now we know that in 1.2 mL of liquid mercury, there are 4.90*10^{22}atoms present.

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7.) A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To
marysya [2.9K]

Answer:

T₂ = 721 k

Explanation:

Given data:

Initial volume = 285 mL

Initial pressure = 1.88 atm

Initial temperature = 355 K

Final temperature = ?

Final volume = 435 mL

Final pressure = 2.50 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂  

T₂  =  P₂V₂ T₁  / P₁V₁

T₂ = 2.50 atm × 435 mL × 355 K / 1.88 atm × 285 mL  

T₂ = 386062.5 atm. mL. K /535.8 atm. mL

T₂ = 721 k

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7 0
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7 0
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