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Allisa [31]
3 years ago
8

Why is there no chemical process that will break down carbon

Chemistry
2 answers:
Jobisdone [24]3 years ago
6 0

There is only a net gain of only two ATP molecules in glycolysis of one six-carbon glucose because it uses the other two to split the glucose.

Veronika [31]3 years ago
5 0

carbon is a strong material

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What part of an atom is involved in a chemical reaction?
Katyanochek1 [597]

Answer:

The answer is supposed to be "Electron cloud" or "Electon".

5 0
4 years ago
what was the purpose of running a tlc of ferrocene, the acetylferrocene product mixture, and co-spot (an overlay of ferrocene an
Mademuasel [1]

Answer:

a. All of the below

Explanation:

Column chromatography in chemistry is a chromatography method used to isolate a single chemical compound from a mixture.

TLC can be used to analyze a chemical reaction to determine if the reactants have been consumed and a new product has formed. Running a tlc of ferrocene, the acetylferrocene product mixture, and co-spot, when you view the TLC plate under a UV light, you will notice that acetylferrocene product is on the right-most lane, this shows that the reaction appears to be a success: the higher spot of ferrocene has been consumed), and a new product spot is present. From these we can observe that the we can use this to identify the number of products of the reaction, determine if the starting material is still present in the reaction and the elution in a chromatography column of the starting material and products. Therefore, all the choices are correct.

8 0
3 years ago
Determine the enthalpy change for the reaction 2C(s) + 2H2O(g) → CH4(g) + CO2(g) using the following:
Sophie [7]

Answer : The enthalpy change for the reaction is, 97.7 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main chemical reaction is,

2C(s)+3H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=131.3kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=41.2kJ

(3) CH_4(g)+H_2O(g)\rightarrow 2H_2(g)+CO(g)    \Delta H_3=206.1kJ

Now we are multiplying reaction 1 by 2 and reversing reaction 3 and then adding all the equations, we get :

(1) 2C(s)+2H_2O(g)\rightarrow 2CO(g)+2H_2(g)     \Delta H_1=2\times 131.3kJ=262.6kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=41.2kJ

(3) 2H_2(g)+CO(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206.1kJ

The expression for enthalpy of main reaction will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(262.6)+(41.2)+(-206.1)

\Delta H=97.7kJ

Therefore, the enthalpy change for the reaction is, 97.7 kJ

3 0
4 years ago
Read 2 more answers
9.Competitive inhibition occurs when a O A. substrate binds to an enzyme in the active site and activates the enzyme B. molecule
alexgriva [62]

Answer:

.Competitive inhibition occurs when a O A. substrate binds to an enzyme in the active site and activates the enzyme B. molecule binds to an enzyme in the active site and prevents the substrate from binding C. molecule binds to an enzyme at a site other than the active site and inhibits the substrate from binding O D. molecule binds to an enzyme but doesn't change the shape of the active site Mark for review (Will be highlighted on the review page) << Previous Question Next Question >>

5 0
3 years ago
If 3.95 g of N2H4 reacts and produces 0.750 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?
adell [148]
Molar mass of N2H4 = 32 grams/mole 
<span>3.95 grams of  N2H4 = 3.95/32
                                  = 0.123 moles </span>

<span>This will produce 0.123 moles of N2 </span>
<span>Now,
From the gas law equation. </span>
<span>P.V = n x R x T </span>
<span>P = 1 atm (given)
V = </span><span>0.123</span><span> x 0.082057 x 295 </span>
<span>V = 2.97 Liters </span>

<span>Theoretical yield = 2.97 Liters.
Actual yield = 0.750 Liters </span>

percentage yield = (0.75/2.97) x 100 %
                           = 25.25 %
3 0
3 years ago
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