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3 years ago

45.6 grams of CH4 is placed into a 275 mL container at 55.00C. What is the pressure inside the container in atm?

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

4 0

Given-

Mass of CH4 - 45.6 grams

Molar mass of CH4 - 16.04 g/mol

volume - 275mL

Temperature- 55°C

Solution,

moles of methane (CH4) n = 45.6/16.04 

= 2.84 moles

Using Formula,

Pv = nRT

P*275 = 2.84*8.314*55

P = 1298.64/275

P = 4.72 Atm.

Answer - The pressure inside the container in atm is 4.72.

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3 years ago

Write the following in standard notation: 5.1874 x 102

wolverine [178]

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2 years ago

Read 2 more answers

A 0.130 mole quantity of NiCl 2 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Ni 2 + ions at equilib

rosijanka [135]

This is an incomplete question, here is a complete question.

A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

Explanation : Given,

Moles of $NiCl_2$ = 0.130 mol

Volume of solution = 1 L

$Concentration=\frac{Moles }{Volume}$

Concentration of $NiCl_2$ = Concentration of $Ni^{2+}$ = 0.130 M

Concentration of $NH_3$ = 1.20 M

$K_f=5.5\times 10^8$

The equilibrium reaction will be:

$Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}$

Initial conc. 0.130 1.20 0

At eqm. x [1.20-6(0.130)] 0.130

= 0.42

The expression for equilibrium constant is:

$K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}$

Now put all the given values in this expression, we get:

$5.5\times 10^8=\frac{(0.130)}{(x)\times (0.42)^6}$

$x=4.31\times 10^{-8}$

Thus, the concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

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4 years ago

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