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STatiana [176]
4 years ago
11

Equations can be balanced by using the half-reaction method. Which step should be completed immediately after finding the oxidat

ion states of atoms?
A;inserting the coefficients
B;balancing the half reactions
C;identifying the half reactions
D;inspecting the number of atoms
Chemistry
2 answers:
Dovator [93]4 years ago
8 0
After finding the oxidation states of atoms, you identify the half reactions (option c).

The half reactions are given by the change of the oxidation states of the atoms.

For example if Cu is in the left side with oxidation state 0 and in the other side with oxidation state 2+, then there you have a half reaction (oxidation reaction). And if you have O with oxidation state 0 in the left side  and with oxidation state 2- in the right side, there you have other half reaction (reducing reaction).
Marat540 [252]4 years ago
5 0

Answer:

c

Explanation:

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3 0
3 years ago
To convert 20 g of ice at -10°C to 110°C to steam you need<br> cal of energy?
shtirl [24]

Answer:

29,200 cal = 1.22 E 5 joules

Explanation:

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6 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
Imagine that you're doing an experiment that requires you to smell the odor of a chemical solution in a beaker. Which procedure
Alex787 [66]
The correct answer is (a) wave the fumes toward your nose with your hand. If you smell the chemicals directly, it could be harmful too your health, especially if they are strong. Also remember to <em>never </em>smell chemicals unless you are being told to do so.
8 0
4 years ago
Read 2 more answers
Find the number of moles of water that can be formed if you have 138 mol of hydrogen gas and 64 mol of oxygen gas.
charle [14.2K]
2H₂₍g₎ + O₂ ₍g₎→ 2H₂O

138 mol H₂ × (2 mol H₂O ÷ 2 mol H₂)= 138 mol H₂O
64 mol O₂ × (2 mol H₂O ÷ 1 mol O₂)= 128 mol H₂O

128 mol H₂O
6 0
4 years ago
Read 2 more answers
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