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klasskru [66]
2 years ago
10

A 7.50 liter sealed jar at 18 degrees C contains 0.125 moles of oxygen and 0,125 moles of nitrogen gas. What is the pressure in

the container?
Chemistry
1 answer:
LiRa [457]2 years ago
6 0
PV = nRT —> P = nRT/V
n = 0.250 moles of gas
R = 0.08206 L atm / mol K
T = 18 + 273 = 291 K
V = 7.50 L

P = (0.250)(0.08206)(291)/(7.50) = 7.96 atm
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Density can be calculated by dividing an object's mass by its volume. If the mass of an object is 10 grams and its volume is 2 m
Luden [163]

Explanation:

  1. mass =10g. volume=2ml. density=mass/volume = 10/2=5kg/m^3
6 0
3 years ago
What is the approximate percentage of nitrogen in<br> the Earth's current atmosphere?
RoseWind [281]

Answer:

78% nitrogen

Explanation:

It has been long agreed upon that our atmosphere is highly abundant with nitrogen. Our atmosphere is composed of  78% nitrogen and then other gases follow such as 21% oxygen, 0.9% argon, and 0.03% carbon dioxide with very small percentages of other elements.

7 0
2 years ago
How can a short-term environmental change, such as a drought or a flood, affect organisms?
viktelen [127]

Answer:

it can affect things by drastically chagning the way that organisms opareate such as the eco systems, the health of the land the flood or drought is on and etc.

Explanation:

hope this helps!

5 0
2 years ago
What is the mole to mole ratio of LiOH to HBr for the following reaction?
umka21 [38]

Explanation:

so moles = mass ÷ mr (1+ 79.9)

so 10.00g ÷ 80.9

which is 0.1236093943

so to 3 S.F is 0.124 moles

also there is 1 to 1 ratio for LiOH to HBr

hope this helps :)

8 0
3 years ago
Dissolution of KOH, ΔHsoln:
swat32

Using Hess's law we found:

1) By <em>adding </em>reaction 10.2 with the <em>reverse </em>of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)   ΔH  (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the <em>total </em>change in enthalpy (ΔH).    

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq)   ΔHsoln    (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)     ΔHneut     (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from <u>differents changes</u> in various <em>reactions </em>can be calculated as the <u>sum</u> of all the <em>enthalpies</em> of all those <em>reactions</em>.      

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq)    (10.3)

We need to <em>add </em>reaction 10.2 to the <u>reverse</u> of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

<u>Canceling</u> the KOH(s) from both sides, we get <em>reaction 10.3</em>:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)    (10.3)

2) The change in enthalpy for <em>reaction 10.3</em> can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

\Delta H = \Delta H_{soln} + \Delta H_{neut}

The enthalpy of <em>reaction 10.1 </em>(ΔHsoln) changed its sign when we reversed reaction 10.1, so:

\Delta H = \Delta H_{neut} - \Delta H_{soln}

Therefore, the ΔHsoln must be <u>subtracted</u> from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

  • brainly.com/question/2082986?referrer=searchResults
  • brainly.com/question/1657608?referrer=searchResults  

I hope it helps you!

6 0
2 years ago
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