Amixture of 10.0 g of ne and 10.0 g ar have a total pressure of 1.6 atm. what is the partial pressure of ne?

1.1atm

Explanation: A mixture of 10.0 g of Ne and 10.0 g of Ar has a total pressure of 1.6 atm. What is the partial pressure of Ne?

a. 1.1 atm

b. 0.8 atm

c. 0.54 atm

d. 0.40 atm

e. 1.3 atm

In any mixture of gases, the partial pressure of the individual components of the mixture is the pressure which that component would normally occupy if it were present alone under the same circumstance. To calculate the partial pressure (p) of Ne in a mixture of Ne and Ar, the partial pressure of a gas in a mixture is given by mole fraction (X) of the gas multiplied by the total pressure (P) of the mixture.

n Ne = 10g /20.18 g/gmol = 0.4955 moles

n Ar = 10g /39.95 g/gmol = 0.2503 moles

Total number of moles = 0.4955 + 0.2503 = 0.7458 moles

X Ne = 0.4955 / 0.7458 = 0.6644

p Ne = (X Ne)(P)

= 0.6644 x 1.6 = 1.063 atm

Which is approximately 1.1atm as answer

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when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol