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Anna35 [415]
3 years ago
8

Which statement about enzymes is true? They lower the free energy of the products. They are not permanently altered by the react

ions they catalyze. Each catalyzes reactions of a broad range of substrates. They are only rarely regulated. They catalyze reactions by raising activation energy.
Chemistry
1 answer:
alexdok [17]3 years ago
8 0

Answer:They are not permanently altered by the reaction they catalyze.

Explanation: Enzymes are usually in lower concentration than substrate molecules they catalyze. Hence an enzyme catalyzes as many substrate molecules as it can. So when an enzyme binds a substrate to it's active site, it does this so as to increase the reaction rate which otherwise would not have been possible without the enzyme. It doesn't mean that the enzyme itself takes part in the chemical reaction. Hence, once an ES(Enzyme-substrate) moves to P(product), the product leaves the active site and the enzyme returns to it's original confirmation ready for binding another molecule of the substrate. Therefore, the enzyme is altered transiently in order to allow the substrate fit into it's active site. Its never altered permanently

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The density of copper is 8.96g/mL. The mass of 7.00 of copper is
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Identify the correct coefificients to balance it<br> -C3H8+O2 to -CO2 +-H2O
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Answer:

{\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

Explanation:

{\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}.

Among the four species in this reaction, \rm C_3H_8 is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is 1.

{\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}.

Number of atoms on the left-hand side of the reaction:

  • \rm C: 1 \times 3 = 3.
  • \rm H: 1 \times 8 = 8.
  • \rm O: not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, \rm CO_2 is the only product with carbon atoms, whereas \rm H_2O is the only product with hydrogen atoms. These 3 carbon atoms and 8 hydrogen atoms would correspond to:

  • 3 / 1 = 3 \rm CO_2 molecules, and
  • 8 / 2 = 4 \rm H_2O molecules.

{\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

Number of atoms on the right-hand side of the reaction:

  • \rm C: 3 \times 1 = 3.
  • \rm H: 4 \times 2 = 8.
  • \rm O: 3 \times 2 + 4 \times 1 = 10.

The number of \rm O atoms on the left-hand side should match those on the right-hand side. In this reaction, \rm O_2 is the only reactant with \rm O\! atoms. These 10 \rm \! O atoms would correspond to:

  • 5 \rm O_2 molecules.

{\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

5 0
2 years ago
When an aqueous solution of magnesium nitrate is mixed with an aqueous solution of potassium carbonate, ____________.?
spin [16.1K]
  <span>Ca(NO3)2 + Na2CO3 = CaCO3 + 2NaNO3 
Yes a precipitate of Calcium Carbonate is formed since it is insoluble in water. 
Mol Wt of Calcium Nitrate is 164. And that of Calcium Carbonate is 100. 
One mole of Calcium Nitrate produces one mole of Calcium Carbonate. 
i.e. 164 gms will produce 100gms of precipitate 
So, 1.74gms of Calcium Carbonate will be obtained from 2.85gms Calcium Nitrate present in the original solution.</span>
3 0
3 years ago
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