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Lostsunrise [7]
3 years ago
5

Industries use heterogeneous catalysts because they are _______

Chemistry
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

Introduction. Heterogeneous catalysts play an important role in industrial chemical production.

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Makovka662 [10]
C for sure 1837462928274628194771
4 0
3 years ago
The galvanic cell illustrated above generates a potential of +1.10V. For the construction of a second galvanic cell (not shown),
xxTIMURxx [149]

Answer:

The initial  E°  for the second cell is the same as for the first cell.

Explanation:

Because the overall chemical reaction that occurs in the cell does not change.

5 0
3 years ago
Calculate the value of the equilibrium constant, Kc, for the reaction below, if 0.208 moles of sulfur dioxide gas, 0.208 moles o
Harman [31]
First, we convert the moles of each substance into the concentration using the volume of the reactor.
[SO₃] = 0.425/1.5 = 0.283 M
[SO₂] = 0.208 / 1.5 = 0.139 M
[O₂] = 0.208/1.5 = 0.139 M
The equilibrium constant is calculated by:
Kc = [SO₃]² / [O₂][SO₂]²
Kc = (0.283)²/(0.139)(0.139)²
Kc = 29.8 = 2.98 x 10¹

The answer is C
8 0
4 years ago
Create an outline to organize your presentation. Write the outline for your presentation in the space provided. You can organize
Alex Ar [27]

Answer:

explain on which topic we have to write and organize an essay ?

Explanation:

5 0
3 years ago
What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0
4 years ago
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