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Romashka [77]
2 years ago
7

How to write a catalyst in a chemical equation

Chemistry
1 answer:
saw5 [17]2 years ago
3 0

Answer: If you write the chemical equation with a right arrow, you can put the name of the catalyst, or its chemical symbol, or some other expression over the arrow.

Explanation: It is what my teachers at college taught me to do, so I just share my experiment to help.

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Which is located inside the lung?
ZanzabumX [31]

Answer:what are the options????

Explanation:

4 0
3 years ago
What is a likely result of an increase in average global temperatures on Earth?
Firdavs [7]

Answer:

A.

Explanation:

Warmer temperatures can also lead to a chain reaction of other changes around the world. That's because increasing air temperature also affects the oceans, weather patterns, snow and ice, and plants and animals. The warmer it gets, the more severe the impacts on people and the environment will be.

8 0
3 years ago
Using the equation, 4Fe + 3O2 Imported Asset 2Fe2O3, if 8 moles of iron and oxygen from the air were available, how many moles o
sergiy2304 [10]
From the balanced chemical equation,
                               4Fe + 3O₂ --> 2Fe₂O₃
it is evident that for every 4 moles of Fe (iron), 2 moles of iron (III) oxide will be formed. Using the same ratio, 
                               8 moles of iron x (2 moles Fe₂O₃ / 4 moles of Fe)
                                         = 4 moles Fe₂O₃
Therefore, the answer is 4 moles. 
3 0
4 years ago
Read 2 more answers
A pH scale reading 13 indicates a ______
lukranit [14]

Answer:

A pH scale reading 13 indicates a strong base.

Explanation:

From my understanding:

1 -4 is a strong acid

4 - 7 is weak acid

7 - 9 is a weak base

9 - 14 is a strong base

6 0
2 years ago
Read 2 more answers
What is the ratio of the concentrations of H2PO4 and HPO4 2 at (a) pH 7.0; (b) pH 7.5; (c) pH 8.0?
mel-nik [20]

Answer:

A. 1.62

B. 0.513

C. 0.162.

Explanation:

pH = pKa + log [ HPO42-] / [H2PO4-]

Assuming, pKa is 7.21

Ka = 6.2 x 10^-8

pKa = 7.21

A.

7.0 = 7.21 + log [HPO42-] / [HPO4-]

- 0.21 = log [ HPO42-] / [H2PO4-]

[ HPO42-] / [ H2PO4-] = 10^-0.21

= 0.617

[H2PO4-] / [HPO42-] = 1 / 0.617

= 1.62

B.

7.5 = 7.21 + log [HPO42-] / [H2PO4-]

0.29 = log [HPO42-] / [H2PO4-]

[HPO42-] / [H2PO4-] = 10^0.29

= 1.95

[H2PO4-] / [HPO42-] = 1/1.95

= 0.513

8.0 = 7.21 + log [HPO42-] / [ H2PO4-]

0.79 = log [HPO42-] / [H2PO4-]

[HPO42-] / [H2PO4-] = 10^0.79

= 6.17

[H2PO4-] / [HPO42-] = 1/6.17

= 0.162

4 0
4 years ago
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