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3 years ago

Put hydrogen bonds dispersion forces and dipole-dipole forces in order

1 answer:

3 0

Increasing order of strength needed to break bonds:
temporary dipole induced dipole interactions
Permanent dipole induced dipole interactions
Hydrogen bonding

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Cylinder A and Cylinder B are sealed, rigid cylinders with movable pistons. Each cylinder contains 500. milliliters of a gas sam

Romashka-Z-Leto [24]

Answer:

The pressure of N₂ gas in cylinder B when compressed at constant temperature increases due to the increase in the frequency of collision between the gas molecules with themselves and with the wall of their container caused by a decrease in volume of the container.

Explanation:

Gas helps to explain the behavior of gases when one or more of either temperature, volume or pressure is varying while the other variables are kept constant.

In the gas cylinder B, the temperature of the given mass of gas is kept constant, however, the volume is decreased by pushing the movable piston farther into the cylinder. According to the gas law by Robert Boyle, the volume of a given mass of gas is inversely proportional to its pressure at constant temperature. This increase in pressure is due to the increase in the frequency of collision between the gas molecules with themselves and with the wall of their container caused by a decrease in volume of the container. As the cylinder becomes smaller, the gas molecules which were spread out further become more packed closely together, therefore, their frequency of collision increases building up pressure in the process.

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3 years ago

Consider the chemical equation for the ionization of CH3NH2 in water. Estimate the percent ionization of CH3NH2 in a 0.050 M CH3

anyanavicka [17]

Answer: The percent ionization of $CH_3NH_2$ in a 0.050 M $CH_3NH_2(aq)$ solution is 8.9 %

Explanation:

$CH_3NH_2+H_2O\rightarrow OH^-+CH_3NH_3^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_b=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= concentration = 0.050 M and $\alpha$ = degree of ionisation = ?

$K_b=4.4\times 10^{-4}$

Putting in the values we get:

$4.4\times 10^{-4}=\frac{(0.050\times \alpha)^2}{(0.050-0.050\times \alpha)}$

$(\alpha)=0.089$

percent ionisation = $0.089\times 100=8.9\%$

8 0

3 years ago

PLEASE HELP!!! Just need an answer, no explanation

Alex

Answer: N2

blah blah blah blah blah blah

6 0

3 years ago

Read 2 more answers

reddi tstudent has a thin copper beaker containing 100 g of a pure metal in the solid state. The metal is at 215°C, its exact me

prohojiy [21]

Explanation:

A point of temperature at which both solid and liquid state of a substance remains in equilibrium without any change in temperature then this temperature is known as melting point.

For example, melting point of water is $0 ^{o}C$ . So, at this temperature solid state of water and liquid state are present in equilibrium with each other.

Therefore, when a 100 g of given pure metal in solid state is heated at its exact melting point which is $215^{o}C$ then some of the solid will change into liquid state but the temperature will remains the same.

4 0

3 years ago

We can consider a chemical to be safe if:

slamgirl [31]

We can consider a chemical to be safe if it does not contain any harmful substances such as nitrogen, harmful acids, or even excessive heat. Other chemicals such as the chemicals used in food products like, for example, citric acid, is not so harmful.
Hope this helps! :D
Any questions? Just let me know! I'd be happy to help any way possible.

8 0

3 years ago