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sveta [45]
3 years ago
13

What are no examples of ozone

Chemistry
1 answer:
Reptile [31]3 years ago
6 0

Answer:

<h2><u>-BLANK-</u> = <em><u>n</u></em>O <em><u>o</u></em>Z<em><u>o</u></em>N<em><u>e</u></em></h2>

<em><u>Explanation:</u></em>

<h2><u><em>strayt fax</em></u></h2>
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For 20 points.... answer this
Allisa [31]
The answer is B.Let it sit
8 0
3 years ago
Read 2 more answers
When will diffusion slow down
scZoUnD [109]

Answer:

Exchange across cell membranes - diffusion. The higher the concentration gradient, the faster the rate of facilitated diffusion, up to a point. As equilibrium is reached the concentration gradient is much lower because the difference in concentrations is very small so the rate of facilitated diffusion will slow down and level off.

Explanation:

6 0
3 years ago
Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. Include all phases in answer
liq [111]
<span><span>Mn<span>O<span>2<span>(s)</span></span></span>+<span>H<span>2<span>(g)</span></span></span>→Mn<span>O<span>(s)</span></span>+<span>H2</span><span>O<span>(g)</span></span></span></span>
4 0
3 years ago
Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.
bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr <em>(1)</em>

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>

<u><em /></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

<em></em>

I hope it helps!

5 0
3 years ago
I need help can someone please do so?
larisa86 [58]

Answer:

0.296 J/g°C

Explanation:

Step 1:

Data obtained from the question.

Mass (M) =35g

Heat Absorbed (Q) = 1606 J

Initial temperature (T1) = 10°C

Final temperature (T2) = 165°C

Change in temperature (ΔT) = T2 – T1 = 165°C – 10°C = 155°C

Specific heat capacity (C) =..?

Step 2:

Determination of the specific heat capacity of iron.

Q = MCΔT

C = Q/MΔT

C = 1606 / (35 x 155)

C = 0.296 J/g°C

Therefore, the specific heat capacity of iron is 0.296 J/g°C

8 0
3 years ago
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