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Vadim26 [7]
3 years ago
12

Given a fixed amount of gas in a rigid container (no change in volume), what pressure will the gas exert if the pressure is init

ially 1.50 atm at 22.0oC, and the temperature is changed to 11.0oC?
A. 301 atm
B. 1.56 atm
C. 0.750 atm
D. 1.44 atm
E. 3.00 atm
Chemistry
1 answer:
Masja [62]3 years ago
8 0

Answer:

The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)

Explanation:

Gay Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move more rapidly. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T}=k

Where P = pressure, T = temperature, K = Constant

You have a gas that is at a pressure P1 and at a temperature T1. When the temperature varies to a new T2 value, then the pressure will change to P2, and then:

\frac{P1}{T1}=\frac{P2}{T2}

In this case:

  • P1= 1.50 atm
  • T1= 22 °C= 295 °K (being 0°C= 273 °K)
  • P2= ?
  • T2= 11 °C= 284 K

Replacing:

\frac{1.5 atm}{295 K}=\frac{P2}{284 K}

Solving:

P2= 284 K*\frac{1.5 atm}{295 K}

P2=1.44 atm

<u><em>The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)</em></u>

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Gala2k [10]

2.2 x 10^-2

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0.00022 * 100% = 0.022% = 2.2 * 10^-2

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Which statement about entropy is correct? - Entropy is sometimes described as the degree of randomness in a system. - solids are
marysya [2.9K]

Answer: Entropy is sometimes described as the degree of randomness in a system.

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2 years ago
How long will it take for a 750 mg sample of radium with a half life of 15 days to decay to exactly 68mg?
weqwewe [10]

Answer:

52 da  

Step-by-step explanation:

Whenever a question asks you, "How long to reach a certain concentration?" or something similar, you must use the appropriate integrated rate law expression.

The i<em>ntegrated rate law for a first-order reaction </em>is  

ln([A₀]/[A] ) = kt

Data:

[A]₀ = 750 mg

 [A] =    68 mg

t_ ½ =   15 da

Step 1. Calculate the value of the rate constant.

 t_½ = ln2/k     Multiply each side by k

kt_½ = ln2         Divide each side by t_½

      k = ln2/t_½

         = ln2/15

         = 0.0462 da⁻¹

Step 2. Calculate the time

ln(750/68) = 0.0462t

         ln11.0 = 0.0462t

            2.40 = 0.0462t     Divide each side by 0.0462

                   t = 52 da

8 0
3 years ago
What is the limiting reactant when 4 mol P4 and 4 mol S8 react.
In-s [12.5K]

Taking into account the reaction stoichiometry, P₄ will be the limiting reagent.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

8 P₄ + 3 S₈ → 8 P₄S₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • P₄: 8 moles
  • S₈: 3 moles
  • P₄S₃: 8 moles

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 3 moles of S₈ reacts with 8 moles of P₄, 4 moles of S₈ reacts with how many moles of P₄?

moles of P_{4} =\frac{4 moles of S_{8} x8 moles of P_{4} }{3 moles of S_{8}}

<u><em>moles of P₄= 10.667 moles</em></u>

But 10.667 moles of P₄ are not available, 4 moles are available. Since you have less amount of moles than you need to react with 4 moles of S₈, P₄ will be the limiting reagent.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

6 0
2 years ago
Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.
bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr <em>(1)</em>

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>

<u><em /></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

<em></em>

I hope it helps!

5 0
3 years ago
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