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3 years ago

Convert 73.7L to KL *

Chemistry

1 answer:

Mariulka [41]3 years ago

7 0

0.737 KL would be the answer, pls give brainliest.

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Which of the following would have similar properties to sodium (Na)?

bogdanovich [222]

Answer:

Explanation:

Elements in the same group (in this case, group 1 or group 1a) have similar chemical properties. The only one in the same group as Sodium (group 1) is Cesium.

A) Neon (Group 8a)(Group 18)

B) Magnesium (Group 2a)(Group 2)

C) Cesium (Group 1a)(Group 1)

D) Phosphorous (Group 5a)(Group 15)

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3 years ago

WHEN IN FRICTION not useful

Paladinen [302]

Friction is not useful in Machinery sometimes, Since it leads to the wear and tear of machine parts.

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3 years ago

A balloon filled with helium gas occupies 2.50 L at 25°C and 1.00 atm. When released, it rises to an altitude where the temperat

stira [4]

Answer:

8,19 L

Explanation:

The combined gas law is a equation that can be used when the initial and final conditions -pressure,volume,amount of moles,temperature-, of a gas change during a process. It can be written:

$\frac{P1V1}{n1T1}=\frac{P2V2}{n2T2}$

If the amount of the gas remains constant, then n1=n2 and we have:

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

For the problem we have:

Note: When working with gases is important to use absolute temperature values (°K, °K=°C+273,15):

P1=1 atm, V1=2,5L, T1=25+273,15=298,15°K

P2=0,3 atm, V2=?, T2=20+273,15=293,15°K

$V2=\frac{P1V1T2}{T1P2}=\frac{1atm*2,5L*293,15K}{298,15K*0,3atm}=8,19L$

The new volume of the balloon is 8,19 L.

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3 years ago

A certain weak acid, HA, has a Ka value of 2.6×10−7. Calculate the percent ionization of HA in a 0.10 M solution.

Andrej [43]

Answer:

The percent ionization is 0,16%

Explanation:

The percent ionization is defined as the number of ions that exist in a substance.

$PI=\frac{[A-]}{[HA]} x100$

First, we find the [A-] using the ka equation

HA ⇄ $H^{+} + A^{-}$

[H+] = [A-]

$Ka=\frac{[H+][A-]}{[HA]}\\ \\$

since the ionization constant is very small we can assume that the final concentration of [HA] is the same

$Ka=\frac{[H+]^{2} }{[HA]} \\\\$

$[H+]=\sqrt[2]{Ka.[HA]} \\\\$

$[H+] =\sqrt{(2,610^{-7} )(0,1)} = 1,61210^{-4}$

Now we calculate the percent ionization using these values

$PI=\frac{1,61210^{-4} }{0,1} X100$

PI=0,16%

4 0

4 years ago

Now imagine that you're waiting for your soup to cool down, and you leave it on the counter while you go do your homework. Your

Vlad [161]

Explanation:

Generally, heat flows from a hot environment to a cold (lesser temperature) environment. In this case, the soup is the hot environment and the air is the cold temperature.

Heat would continue to flow from one environment to another until thermal equilibrium is reached. At this thermal equilibrium, both environments would have the same temperature.

4 0

3 years ago