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Arturiano [62]
3 years ago
6

The pressure of a gas is 1.0 atm, the volume is 3.0 L, and the temperature is 200 K. A chemist changes one factor while keeping

another constant so that the new pressure is 2.0 atm. Which of the following could be the new conditions?
The final volume is 6.0 L, while temperature is kept constant.
The final temperature is 100 K, while volume is kept constant.
The final volume is 4.0 L, while temperature is kept constant.
The final temperature is 400 K, while volume is kept constant.
Chemistry
1 answer:
anygoal [31]3 years ago
4 0

Answer: fourth statement is correct. The final temperature is 400 K, while volume is kept constant.

Explanation: At constant temperature, volume is inversely proportional to the pressure and,

At constant volume, pressure is directly proportional to the kelvin temperature.

Initially the gas pressure is 1.0 atm, volume is 3.0 L and the temperature is 200 K.  The new pressure is 2.0 atm.

If we consider the temperature is constant, then new volume must be 1.5 L since the pressure is doubled and we know that volume is inversely proportional to the pressure.

So, as the pressure is doubled, the volume will be halved means it will decrease from 3.0 L to 1.5 L.

First statement is not correct since it says the new volume will be 6.0 L while temperature is kept constant.

If the volume is kept constant, the temperature must be doubled as the pressure has also doubled. So, new temperature will be 400 K.

second statement is also not correct as it says, the final temperature is 100 K, while volume is kept constant.

Third statement is also not correct as we already have seen that the new volume will be 1.5 L, while the temperature is kept constant.

Hence, the only and only correct statement is fourth as the final temperature will be 400 K(double of initial 200 K), while the volume is kept constant.

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Answer:

ii. C,\ N^-,\ O^{2-}

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From the given options:-

C,\ N^-,\ O^{2-} are the isoelectric species.

All the three species have same number of electrons which is equal to 6.

Carbon has 6 electrons.

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What type of reaction is represented by the following? A + B ⟶ AB
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2 years ago
Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction
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Answer:

<u>Balanced equation:</u>

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

<u>Ionic equation:</u>

Pb^{2+}(aq)+2NO_{3}^{-}(aq)+2K^{+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)+K^{+}(aq)+2NO_{3}^{-}

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

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g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc
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\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}

\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}

\text{and in the liquid form it is easily transported. An industrial chemist studying this}

\text{reaction fills a} \ \mathbf{100 \  L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \  \text{of oxygen gas, }

\text{to be} \  \mathbf{2.6\  mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}

\text{ammonia at the final temperature of the mixture. Round your answer to  2 significant digits.}

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

At equilibrium;

The water vapor = \dfrac{2.6 \ mol}{100 \ L} = 6x

x = \dfrac{2.6}{100} \times \dfrac{1}{6}

x = 0.00433

\text{equilibrium constant}  ({k_c}) =  \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }

\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\

Replacing the value of x, we have:

K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}

K_c = \mathbf{5.5 \times 10^{-8} \ to  \ 2 \ significant \ figures}

5 0
3 years ago
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