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Otrada [13]
3 years ago
5

Suppose you plant a seed and observe that a tree of large mass grows from it. The tree achieves a final mass that changes very l

ittle for years afterward. Which of the following is true about the tree?
A. Both anabolic and catabolic reactions took place in the seed and tree when it was young and growing, but then all reactions stopped when the tree reached a stable mass.
B. Both anabolic and catabolic reactions took place in the seed and tree when it was young and growing, and both continue now even though the tree reached a stable mass.
C.Only anabolic reactions
D.Only catabolic reactions
Chemistry
2 answers:
Marta_Voda [28]3 years ago
5 0
<span>Suppose you plant a seed and observe that a tree of large mass grows from it. The tree achieves a final mass that changes very little for years afterward.<span>

</span>Answer: Of the options presented above the one that is true about the tree is answer choice B) Both anabolic and catabolic reactions took place in the seed and tree when it was young and growing, and both continue now even though the tree reached a stable mass.

I hope it helps, Regards.</span>
Vanyuwa [196]3 years ago
5 0

Answer is: B. Both anabolic and catabolic reactions took place in the seed and tree when it was young and growing, and both continue now even though the tree reached a stable mass.

Food is broken in series of reactions called catabolism.

Catabolism (catabolic reactions) is the group of metabolic pathways that breaks down molecules of food (for example polysaccharides, lipids, proteins) into smaller units that are either oxidized to release energy or used in other anabolic reactions.

Anabolism (anabolic reactions) is the set of metabolic pathways that construct molecules from smaller units.

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William measures a test tube and finds that the mass of the test tube is 5g. He places the lone reactant in the test tube and fi
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86 percent is the percent yield for this experiment if he expected to produce 5g of product.

Explanation:

Given that:

mass of test tube = 5 grams

mass of test tube + reactant is 12.5 grams

mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)

mass of reactant = 12.5 -5

                             = 7.5 grams

when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.

so mass of product formed = 9.3 - 5

                                             = 4. 3 grams of product is formed (actual yield)

However, he expected the product to be 5 grams (theoretical yield)

Percent yield = \frac{actual yield}{theoretical yield} x 100

          putting the values in the formula:

percent yield = \frac{4.3}{5} x 100

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86 percent is the percent yield.

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Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a
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The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound            Amount

     Ni                        12.7 g

   CO                        1.98 g

Ni(CO)_4                  0.597 g

Calculate the value of the equilibrium constant.

<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1

<u>Explanation:</u>

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of Ni(CO)_4 = 0.597 g

Volume of container = 2.6 L

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M

\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M

\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M

For the given chemical reaction:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

The expression of equilibrium constant for the reaction:

K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1

Hence, the value of equilibrium constant for the reaction is 2448.1

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