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Ksenya-84 [330]
2 years ago
10

When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, pr

ovided that
A) it is a reactant in one intermediate reaction and a catalyst in the other reaction.
B) it is a product in one intermediate reaction and a catalyst in the other reaction.
C) it is a reactant in one intermediate reaction and a product in the other reaction.
D) it is a reactant in both of the intermediate reactions.
Chemistry
2 answers:
Gnesinka [82]2 years ago
4 0
The correct answer in here is <span>it is a reactant in one intermediate reaction and a product in the other reaction. So your option is C. This one explains itself perfectly. I hope this can help you a lot </span>
aalyn [17]2 years ago
3 0

<u>Answer:</u> The correct answer is Option C.

<u>Explanation:</u>

When two intermediate chemical equations combine, the same substance which appears in the same phase on the reactant side can be cancelled out only when it also appears in the same phase on the product side.

<u>For example:</u> The conversion of diamond to graphite follows two intermediate steps, which are written as:

C(s)_{diamond}+O_2(g)\rightarrow CO_2(g)

CO_2(g)\rightarrow C(s)_{graphite}+O_2(g)

As, carbon dioxide and oxygen gases are present in the same phase and in opposite sides of the chemical reaction. So, it can be cancelled out.

Net reaction follows:  C(s)_{diamond}\rightarrow C(s)_{graphite}

Hence, the correct answer is Option C.

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In aqueous solution, hypobromite ion, BrO-, reacts to produce bromate ion, BrO3 -, and bromide ion, Br-, according to the follow
maw [93]

Answer:

22.73s

Explanation:

The reaction is a second order reaction, we know this by observing the unit of the slope.

rate constant = k = 0.056 M-1s-1

the initial concentration of BrO- [A]o = 0.80 M

time = ?

Final concentration [A]t= one-half of 0.80 M = 0.40M

1 / [A]t = kt + 1 / [A]o

1 / 0.40 = 0.056 * t + 1 / 0.80

t = (2.5 - 1.25) / 0.056

t = 22.73s

5 0
3 years ago
Why is it better to conduct an experiment more than once?
lilavasa [31]

The first reason to repeat experiments is simply to verify results. Different science disciplines have different criteria for determining what good results are. Biological assays, for example must be done in at least triplicate to generate acceptable data. Science is built on the assumption that published experimental protocols are repeatable.


2)      The next reason to repeat experiments is to develop skills necessary to extend established methods and develop new experiments. “Practice make perfect” is true for the concert hall and the chemical laboratory.


3)      Refining experimental observations is another reason to repeat. Maybe you did not follow the progress of the reaction like you should have.


4)      Another reason to repeat experiments is to study and/or improve them in way. In the synthetic chemistry laboratory, for example, there is always a desire to improve the yield of a synthetic step. Will certain changes in the experimental conditions lead to a better yield? The only way to find out is to try it! The scientific method informs us that it is best to only make one change at a time.


5)  The final reason to repeat an extraction, chromatographic or synthetic protocol is to produce more of your target substance. This is sometimes referred to scale-up.

8 0
2 years ago
32P can be used to make any nucleotide (A, C, G, or T) radioactive. Which of the following explains why this is true?
xxMikexx [17]
B. The answer is: All nucleotides have a phosphorus atom that can be replaced with 32P.

Nucleotides contain a nitrogenous base, a five-carbon sugar, and, at least, one phosphate group. Exactly that phosphate group in the nucleotide has the phosphorus atom. Therefore, the phosphorus atom in the nucleotide can be replaced with radioactive phosphorus-32 (32P). 
3 0
2 years ago
If 3.91 moles of H2 were used, how many moles of NH3 would be made?<br><br> HELP PLEASE ASAP
matrenka [14]

N + H2 → NH3

balance the equation

2N + 3H2 → 2NH3

3H2 → 2NH3

3 : 2

3.91 ÷ 3 = 1.303 mol

1.3 × 2 = 2.606 mol

7 0
2 years ago
The sample concentration was measured at 50mg/ml. The loading concentration needs to be 10mg/ml. The final volume needs to be 25
Svet_ta [14]

Answer:

a)  V_1=5ul

b)  v=20ul

Explanation:

From the question we are told that:

initial Concentration C_1=50mg/ml

Final Concentration C_2=10mg/ml

Final volume needs V_2 =25ul

Generally the equation for Volume is mathematically given by

C_1V_1=C_2V_2

V_1=\frac{C_1V_1}{C_2}

V_1=\frac{10*25}{50}

V_1=5ul

Therefore

The volume of buffer needed is

v=V_2-V_1\\\\v=25-5

v=20ul

3 0
3 years ago
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