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KiRa [710]
3 years ago
14

Water is placed outside at 298 K overnight. Which statement best describes what would happen?a)298 K converts to -24.9 °C, so th

e water would freezeb)298 K converts to 24.9 °C, so the water would remain in its liquid statec)298 K converts to 24.9 °C, so the water would freeze
Chemistry
1 answer:
user100 [1]3 years ago
7 0

Answer: b)298 K converts to 24.9 °C, so the water would remain in its liquid state

Explanation: Kelvin is an absolute temperature scale, that means that 0K (zero Kelvin) is the lowest temperature possible and compare to Celsius scale the change of 1 degree is the same, but 0°C is the same as 273,1K

So, in order to convert Kelvin to Celsius you have to subtract 273,1

In this case, 298K - 273,1 = 24,9°C

This temperature is room temperature, so water is in liquid state.

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A substance has a solubility of 350 ppm.how many grams of the substance are present in 1.0l of a saturated solution
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Since the given solubility is 350 ppm, convert it first with fraction of solubility. by dividing the solubility with 10^6
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What is the poH of a<br> 2.6 x 10-6 M H+ solution?
andriy [413]

Answer:

Explanation:

pH and pOH.....

The pH is a way of expressing the hydrogen ion concentration.

pH = -log[H+] ............. where [x] means "the concentration of x in moles per liter."

From pH you can compute pOH since at 25C pH + pOH = 14.00 .......... (but only at 25C)

pH = -log(2.6x10^-6) = 5.585 ..... which should be rounded to two significant digits: pH = 5.59

When taking the log of a number, only the digits to the right of the decimal reflect the precision in the original number. Since 2.6x10^-6 has two significant digits, a pH of 5.59 has two significant digits.

pOH + pH = 14.00

pOH = 14.00 - pH = 14.00 - 5.59 = 8.41 ......... at 25C

We can also use the H+ ion concentration to get the hydroxide ion concentration and from that the pOH.

Kw = [H+][OH-] = 1.00x10^-14 .......... at 25C .... like any Kc, the value changes with temperature

[OH-] = Kw / [H+] = 1.00x10^-14 / 2.6x10^-6 = 3.846x10^-9 .... to a couple of guard digits

pOH = -log[OH-] = -log(3.846x10^-9) = 8.415 ...... round to two significant digits: pOH = 8.42 ..... at 25C

=========

Just for grins, you might want to know how Kw changes with temperature, and how [H+] and [OH-] are related at some other temperatures. The pH is the pH of a neutral solution at various temperatures. For instance at 10C a neutral solution has a pH of 7.27. That's not a basic pH. 7.27 is the pH of a neutral solution, but at a different temperature. In a neutral solution at 10C [H+] = [OH-] = 5.41x10^-8M.

pH and Kw for a neutral solution at different temperatures

T .........pH ......... Kw

0......... 7.47....... 0.114 x 10-14

10....... 7.27....... 0.293 x 10-14

20....... 7.08....... 0.681 x 10-14

25....... 7.00....... 1.008 x 10-14

30....... 6.92....... 1.471 x 10-14

40....... 6.77....... 2.916 x 10-14

50....... 6.63....... 5.476 x 10-14

100..... 6.14....... 51.3 x 10-14

4 0
3 years ago
What is the mole fraction of ethanol in a solution of 3.00 moles of ethanol and 5.00 moles of water?
Rus_ich [418]
If theres a mixture of components we can calculate the mole fraction
mole fraction can be calculated as follows
mole fraction of component = \frac{number of moles of component}{total number of moles of all components }
number of moles of ethanol - 3.00 mol
total number of moles in mixture - 3.00 + 5.00 = 8.00 mol 

mole fraction of ethanol = \frac{3.00 mol}{8.00 mol }
mole fraction of ethanol is 0.375
6 0
3 years ago
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