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aliina [53]
3 years ago
6

In the equation 2 H2 + O2 -> 2H2O, H2, O2, and H2O are all trace elements. Only H2O is a compound. Only H2 and O2 are compoun

ds. H2, O2, and H2O are ALL elements. H2, O2, and H2O are ALL compounds.
Chemistry
1 answer:
stiks02 [169]3 years ago
6 0

Answer: only H2O is a compound

Explanation:

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sleet_krkn [62]

Answer:

your answer is 3

Explanation:

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3 years ago
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If the a of a monoprotic weak acid is 2. 6×10−6, what is the ph of a 0. 33 m solution of this acid?
yawa3891 [41]

The pH of the monoprotic weak acid is 2.79.

<h3>What are weak acids?</h3>

The weak acids are the acids that do not fully dissociate into ions in the solution. Strong acids fully dissociate into ions.

The chemical reaction is HA(aq) ⇄ A⁻(aq) + H⁺(aq).

c (monoprotic acid) = 0.33 M.

Ka = 1.2·10⁻⁶

[A⁻] = [H⁺] = x

[HA] = 0.33 M - x

Ka = [A⁻]·[H⁺] / [HA]

2. 6 × 10⁻⁶ = x² / (0.33 M - x)

Solve quadratic equation: [H⁺] = 0.000524 M.

pH = -log[H⁺]

pH = -log(0.000524 M)

pH = 2.79

Thus,  the pH of the monoprotic weak acid is 2.79

To learn more about weak acids, refer to the below link:

brainly.com/question/13032224

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5 0
2 years ago
The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
5 0
3 years ago
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egoroff_w [7]
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What must be satisfied for a hypothesis to be useful?
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Answer:

it must be testable I think that's the answer

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