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2 years ago

11

An ideal gas is contained in a cylinder with a volume of 5.0x102 mL at a temperature of 30°C and a pressure of 710. Torr. The ga

s is then compressed to a volume of 25 mL, and the temperature is raised to 820° C. What is the new pressure of the gas?

1 answer:

Scorpion4ik [409]2 years ago

4 0

Answer:

51207 torr is the new pressure of the gas

Explanation:

We can solve this question using combined gas law that states:

P1V1T2 = P2V2T1

<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of the gas</em>

<em> </em>

Computing the values of the problem:

P1 = 710torr

V1 = 5.0x10²mL

T1 = 273.15 + 30°C = 303.15K

P2 = ?

V2 = 25mL

T2 = 273.15 + 820°C = 1093.15K

Replacing:

710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K

3.881x10⁸torr*mL*K = P2 * 7.579x10³mL*K

P2 = 51207 torr is the new pressure of the gas

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If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the

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Answer:

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Explanation:

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2 years ago

Select True or False: Hydrogen plays an important role in many industrial processes. The following correctly represents a balanc

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Answer: False

Explanation:

Since the given equation is not balanced properly.

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2 years ago

A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de

zhannawk [14.2K]

Answer:

$\boxed{\text{2.6 kPa}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³; T₁ = -10 °C

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Calculations:

(a) Convert temperature to kelvins

T₁ = (-10 + 273.15) K = 263.15 K

(b) Calculate the pressure

$\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}$

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2 years ago

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

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<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

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$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

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3 years ago