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2 years ago

C. How many moles of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg?

1 answer:

5 0

Answer: There are $4.45\times 10^{-4}moles$ of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 895 mm Hg= 1.18 atm (760 mm Hg= 1 atm)

V = Volume of gas = 9.55 ml = 0.00955 L (1 L=1000ml)

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $35^0C=(35+273)K=308K$

$n=\frac{PV}{RT}$

$n=\frac{1.18atm\times 0.00955L}{0.0821L atm/K mol\times 308K}=4.46\times 10^{-4}moles$

Thus there are $4.45\times 10^{-4}moles$ of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg

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Why does a can of diet cola float in water but a can of regular cola does not

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An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h

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Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base = w

Height of the pyramid = h

Volume of the pyramid = $V=\frac{1}{3}lwh$

We have:

Rate at which water is filled in cube = $\frac{dV}{dt}= 45 cm^3/s$

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

$V=\frac{1}{3}\times l^2\times h$

$\frac{l}{h}=\frac{6}{13}$

$l=\frac{6h}{13}$

$V=\frac{1}{3}\times (\frac{6h}{13})^2\times h$

$V=\frac{12}{169}h^3$

Differentiating V with respect to dt:

$\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}$

$\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}$

Putting, h = 4 cm

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}$

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13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

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3 years ago

A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

kaheart [24]

The equilibrium constant of the reaction is 282. Option D

<h3>What is equilibrium constant?</h3>

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

Concentration of iodine = 1.600 mol/1.000-L = 1.600M

In this case, we must set up the ICE table as shown;

Br2(g) + I2(g) ↔ 2IBr(g)

I 0.6 1.6 0

C -x -x +2x

E 0.6 - x 1.6 - x 1.190

If 2x = 1.190

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The concentrations at equilibrium are;

[Br2] = 0.6 - 0.595 = 0.005

[I2] = 1.6 - 0.595 = 1.005

Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

Kc = 282

Learn more about equilibrium constant:brainly.com/question/15118952

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1 year ago

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