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3 years ago

Maintaining constant pressure, the volume of a gas is increased from 12,0 L

Chemistry

1 answer:

kow [346]3 years ago

3 0

Answer:

790 K

Explanation:

Step 1: Given data

Initial volume of the gas (V₁): 12.0 L

Initial temperature of the gas (T₁): 23.0 °C

Final volume of the gas (V₂): 32.0 L

Step 2: Convert 23.0 °C to Kelvin

We will use the following expression.

K = °C + 273.15 = 23.0 + 273.15 = 296.2 K

Step 3: Calculate the final temperature of the gas

Assuming constant pressure and ideal behavior, we can calculate the final temperature of the gas using Charles' law.

T₁/V₁ = T₂/V₂

T₂ = T₁ × V₂/V₁

T₂ = 296.2 K × 32.0 L/12.0 L = 790 K

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The Calvin cycle and the Krebs cycle are cycles of chemical reactions because they

ki77a [65]

Answer: The correct answer is option (b).

Explanation:

A cycle in science is defined as process or phenomena which reoccurs again after a certain interval of a time.

In a set of chemical reactions where reactant is regenerated as a product in the end. This regenerated reactant will again establish the same set of chemical reactions in a same the fashion continuously.

Krebs cycle: it is cycle which occurs during in cellular respiration which starts from acetyl-Co-A enzyme. In the end of cycle energy is released with regeneration of acetyl-Co-A enzyme which will further continue the cycle.

Calvin cycle: It is cycle which occurs in chloroplast during the process of photosynthesis which starts from carbon-dioxide. In the end sugar molecule is released with regeneration of carbon-dioxide molecule which will further continue the cycle.

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3 years ago

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What are the favored geometrical arrangements for abn molecules for which the a atom has 2?

xxMikexx [17]

I believe that the answer to the question provided above is that the <span>favored geometrical arrangements for abn molecules for which the a atom has 2 is in a triangular form.</span>
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

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2 years ago

A solution of Na2CO3 is added dropwise to a solution that contains 1.15×10−2 M Fe2+ and 0.58×10−2 M Cd2+. What concentration of

castortr0y [4]

The question is incomplete, complete question is;

A solution of $Na_2CO_3$ is added dropwise to a solution that contains $1.15\times 10^{-2} M$ of $Fe^{2+}$ and $0.58\times 10^{-2} M$ and $Cd^{2+}$ .

What concentration of $CO_3^{2-}$ is need to initiate precipitation? Neglect any volume changes during the addition.

$K_{sp}$ value $FeCO_3: 2.10\times 10^{-11}$

$K_{sp}$ value $CdCO_3: 1.80\times 10^{-14}$

What concentration of $CO_3^{2-}$ is need to initiate precipitation of the first ion.

Answer:

Cadmium carbonate will precipitate out first.

Concentration of $CO_3^{2-}$ is need to initiate precipitation of the cadmium (II) ion is $3.103\times 10^{-12} M$ .

Explanation:

1) $FeCO_3\rightleftharpoons Fe^{2+}+CO_3^{2-}$

The expression of an solubility product of iron(II) carbonate :

$K_{sp}=[Fe^{2+}][CO_3^{2-}]$

$2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}$

$[CO_3^{2-}]=1.826\times 10^{-9}M$

2) $CdCO_3\rightleftharpoons Cd^{2+}+CO_3^{2-}$

The expression of an solubility product of cadmium(II) carbonate :

$K_{sp}=[Cd^{2+}][CO_3^{2-}]$

$1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}$

$[CO_3^{2-}]=3.103\times 10^{-12} M$

On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.

So, cadmium carbonate will precipitate out first.

And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the $3.103\times 10^{-12} M$ concentration.

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