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nordsb [41]
3 years ago
11

What does is so used to eliminate a substance that is too large to leave by diffusion

Chemistry
1 answer:
dybincka [34]3 years ago
8 0
Reword it so we can understand it better, to answer your quetion

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Why does potassium explode when exposed to water?
soldi70 [24.7K]

Answer:

The highly unstable pure sodium or potassium wants to lose an electron and this splits the water atom, producing a negatively charged hydroxide ion and hydrogen and forming an explosive gas that ignites.

Explanation:

4 0
3 years ago
Answer with explanation
creativ13 [48]
The answer should be:

KOH (aq) + HCl (aq) --> KCl (aq) + H20 (l).

KOH is a base, because OH can accept a H+.
HCl is an acid because it can donate a H+.

In general, bases are : OH-, and acids are : H+.


8 0
3 years ago
60 POINTS! PLEASE ANSWER!
Mumz [18]
I would have helped but I didn’t understand it sorry that I didn’t answer :(
6 0
2 years ago
An unknown substance has a mass of 14.7 g . When the substance absorbs 1.323×102 J of heat, the temperature of the substance is
Fudgin [204]
The correct answer of the given question above would be option B. IRON 0.449. Based on the given details above about an unknown substance that has a mass of 14.7 g and  the substance absorbs 1.323×102 J of heat, the temperature of the substance is raised from 25.0 ∘C to45.0 ∘C, most likely, the substance is IRON. Hope this answers the question.
7 0
3 years ago
Read 2 more answers
The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k
Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})

We have given in the question

P_1=102\ mmHg

T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol

And R is the Universal Gas Constant.

R=0.008 314 kJ/Kmol

ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165

Taking inverse log both side we get,

\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg

8 0
3 years ago
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