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musickatia [10]
2 years ago
11

Help me with this question I’m kind of struggling

Chemistry
2 answers:
maks197457 [2]2 years ago
8 0

Answer:

The answer is A. Humans eat other organisms to get the nutrients they need to survive.

Explanation:

Heterotroph means that organisms need to eat organisms to live.

Hope I could help. If you need any more help ill try my best to provide assistance.

larisa86 [58]2 years ago
3 0

All i know that we are HOOman

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Solar energy.

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Which of these is an example of matter?
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Fe(s) + O2 (g) → Fe3+ O2− What most likely happens during this reaction?
Musya8 [376]

Answer : The most likely happens during this reaction is, Oxidation-reduction

Explanation :

The balanced reaction will be,

4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)

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When iron react with oxygen gas to give iron oxide. This process is known as iron rusting. During the reaction, oxidation-reduction process occurs.

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5 0
3 years ago
Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec
Svetach [21]

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})

The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









4 0
3 years ago
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