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vladimir1956 [14]
4 years ago
12

HELP ME SOLVE THIS PLEASE

Chemistry
2 answers:
tigry1 [53]4 years ago
8 0

Answer:

B. 47.3

Explanation:

Kitty [74]4 years ago
3 0
Option 2: 47.3 mL is the answer!!
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A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in
In-s [12.5K]

Answer:

3.6667

Explanation:

<u>For helium gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{5.6}\times {3.0}={P_2}\times {9.0} atm

{P_2}=\frac {{5.6}\times {3.0}}{9.0} atm

{P_1}=1.8667\ atm

<u>The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm</u>

<u>For Neon gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{3.6}\times {4.5}={P_2}\times {9.0} atm

{P_2}=\frac {{3.6}\times {4.5}}{9.0} atm

{P_1}=1.8\ atm

<u>The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm</u>

<u>Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.</u>

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