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4 years ago

To squeeze a gas into a smaller space word

Chemistry

2 answers:

ElenaW [278]4 years ago

8 0

compress
or compressing gas

hichkok12 [17]4 years ago

7 0

Compress because itis another word for making smaller or putting a big thing in a small place

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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 ME WITH THISSSS!!!

Lilit [14]

16) you can tell which craters are younger by them being small and you can tell which craters are older by them being larger.

17) The larger craters experienced energetic impacts because they are simply larger in size which makes them being impacted.

Hope this was helpful have a nice day :)

4 0

3 years ago

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$