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Allisa [31]
2 years ago
6

What formula was used to find the answer

Chemistry
1 answer:
Setler79 [48]2 years ago
6 0
Can you give more information?
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Are enzyme-catalyzed reactions examples of homo-<br> geneous or heterogeneous catalysis? Explain.
Tanzania [10]

Answer:Yes,enzymes are catalyzed reactions

Explanation:Enzymes are protein that speeds up chemical reactions. Enzyme catalyzed reaction are divided into two:

Homogeneous reaction

Heterogeneous reaction.

Homogeneous catalysts occupy the same phase as the reaction mixture, while heterogeneous catalysts occupy a different phase.

Acid catalysis, organometallic catalysis, and enzymatic catalysis are examples of homogeneous catalysis.

Vanadium oxide (V2 O5) is a brown/yellow solid on which the oxygen and sulfur dioxide can adsorb in order to react with each other to form sulfuric acid.

8 0
3 years ago
Explain why the melting point of calcium is high?
DanielleElmas [232]
It takes more energy to breakdown the bonds
5 0
2 years ago
Read 2 more answers
Please Balanced this Equation
Scilla [17]

Answer:

\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O.

Explanation:

Consider the oxidation state on each of the element:

Left-hand side:

  • O: -2 (as in most compounds);
  • Cr: \displaystyle \frac{1}{2}(\underbrace{-2}_{\text{ion}} - \underbrace{7\times (-2)}_{\text{Oxygen}}) = +6;
  • Fe: +2 (from the charge of the ion);

Right-hand side:

  • Cr: +3;
  • Fe: +3.

Change in oxidation state:

  • Each Cr atom: decreases by 3 (reduction).
  • Each Fe atom: increases by 1 (oxidation).

Changes in oxidation states shall balance each other in redox reactions. Thus, for each Cr atom on the left-hand side, there need to be three Fe atoms.

Assume that the coefficient of the most complex species \rm Cr_2O_7^{2-} is 1. There will be two Cr atoms and hence six Fe atoms on the left-hand side. Additionally, there are going to be seven O atoms.

Atoms are conserved in chemical reactions. As a result, the right-hand side of this equation will contain

  • two Cr atoms,
  • six Fe atoms, and
  • seven O atoms.

O atoms seldom appear among the products in acidic environments; they rapidly combine with \rm H^{+} ions to produce water \rm H_2O. Seven O atoms will make seven water molecules. That's fourteen H atoms and hence fourteen \rm H^{+} ions on the product side of this equation. Hence the balanced equation. Double check to ensure that the charges on the ions also balance.

\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O.

7 0
3 years ago
How much mass (in grams) is in a 3.25 mol sample of NH4OH?
Sergeeva-Olga [200]

Answer:

114 grams

Explanation:

1. NH4OH molecular weight: The molar mass of NH4OH = 35.0458 g/mol...

2. Convert grams-NH4OH-to-moles or moles-NH4OH-to-grams. Molecular weight calculation: 14.0067 + 1.00794*4 + 15.9994 + 1.00794.

3 0
2 years ago
A mixture initially contains A, B, and C in the following concentrations: [A] = 0.650 M, [B] = 1.35 M, and [C] = 0.300 M. The fo
dimaraw [331]

<u>Answer:</u> The value of K_c for given reaction is 0.465

<u>Explanation:</u>

We are given:

Initial concentration of A = 0.650 M

Initial concentration of B = 1.35 M

Initial concentration of C = 0.300 M

Equilibrium concentration of A = 0.550 M

Equilibrium concentration of B = 0.400 M

For the given chemical equation:

                           A+2B\rightarrow C

<u>Initial:</u>                0.65     1.35     0.30

<u>At eqllm:</u>        0.65-x   1.35-2x   0.30+x

Evaluating the value of 'x'

0.650-x=0.550\\\\x=0.100

So, equilibrium concentration of B = 1.35 - 2x = [1.35 - 2(0.100)] = 1.15 M

Equilibrium concentration of C = (0.30 + x) = (0.300 + 0.100) = 0.400 M

The expression of K_c for above equation follows:

K_c=\frac{[C]}{[A][B]^2}

Putting values in above equation, we get:

K_c=\frac{0.400}{0.650\times (1.15)^2}\\\\K_c=0.465

Hence, the value of K_c for given reaction is 0.465

7 0
3 years ago
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