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Otrada [13]
3 years ago
10

Attempt 1 of 1

Chemistry
1 answer:
ra1l [238]3 years ago
6 0

I'm not sure, i had this same question, and it doesn't make any logical sense because

A centimeter is a centimeter but in the video apparently not... for example in the video it showed the notebook as 40cm for one and the next one shows it as 14.4 cm it stayed as cm both times the size of the ruler should not change that one ruler was by 10cm the other was by 1 cm and it showed 2 completely different numbers

so therefore it doesn't make any logical sense

here was my measurements from the little video it had

try getting ahold of the teacher or whoever gave you the question its all you really can do...

hope this helped somewhat even through it probably didn't...

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The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
Anuta_ua [19.1K]

Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

3 0
3 years ago
If hydrochloric acid has a [H+] of 1.2 x 10-2 M, what is the pH?<br> 1.9<br> 0 1<br> 1.2<br> O<br> 8
Vanyuwa [196]

Answer:

<h2>1.9</h2>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ {H}^{+} ]

From the question we have

pH =  -  log(1.2 \times  {10}^{ - 2} )  \\  = 1.920818...

We have the final answer as

<h3>1.9</h3>

Hope this helps you

7 0
3 years ago
Read 2 more answers
Please help!! will give brainliest if u tell me how
Alona [7]
I am pretty sure it’s A
8 0
3 years ago
A 4.0 L container holds a sample of hydrogen gas at 306 K and 150 kPa. If the pressure increases to 300 kPa and the volume remai
riadik2000 [5.3K]

Answer:

612 K

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 306 K

Initial pressure (P₁) = 150 kPa

Final pressure (P₂) = 300 kPa

Volume = 4 L = constant

Final temperature (T₂) =?

Since the volume is constant, the final (i.e the new) temperature of the gas can be obtained as follow:

P₁ / T₁ = P₂ / T₂

150 / 306 = 300 / T₂

Cross multiply

150 × T₂ = 306 × 300

150 × T₂ = 91800

Divide both side by 150

T₂ = 91800 / 150

T₂ = 612 K

Thus, the new temperature of the gas is 612 K

4 0
3 years ago
When objects touch each other, charge can be transferred by
Dovator [93]

Answer:

friction

Explanation:

electrons from one uncharged object to another uncharged object by rubbing. When two uncharged objects rub together, some electrons from one object can move onto the other object. hope this is your right

and not just a riddle

7 0
3 years ago
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