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3 years ago

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Eleven molecules have speeds 15, 16, 17, . . . , 25 m/s. Calculate (a) vavg, and (b) vrms. Part a express your answer with the a

ppropriate units.

2 answers:

lora16 [44]3 years ago

7 0

The speeds are

15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 m/s.

Vavg = 15 + 16 + 17 + 18 + 19 + 20 +21 +22+ 23+24+ 25 / total molecules = 20 m/s

Vrms = ((sum of square of speed of molecules ) / total molecules )^1/2

Vrms = ( (15^2 + 16^2 + 17^2 + ....25^2) / 11 )^1/2

= (225+ 256 + 289 + 324 + 361+ 400 + 441 + 484 + 529 +576 +625 / 11)^1/2

= (4510 / 11 )^1/2 = 20.25 m /s

BlackZzzverrR [31]3 years ago

5 0

Answer:

Average velocity of molecules is 20 m/s and root mean square velocity of molecules is 20.24 m/s

Explanation:

It is given that, eleven molecules have speeds 15,16,17,18,19,20,21,22,23,24,25 m/s. We have to calculate (1) average velocity (2) root mean square velocity of molecules.

Average velocity can be calculated as :

$v_{avg}=\dfrac{sum\ of\ speeds}{no\ of\ molecules}$

$v_{avg}=\dfrac{15+16+17+18+19+20+21+22+23+24+25}{11}$

$v_{avg}=20\ m/s$

Root mean square velocity is given by :

$v_{rms}=\sqrt{\dfrac{sum\ of\ squares\ of\ speeds}{no\ of\ molecules}$

$v_{rms}=\sqrt{\dfrac{15^2+16^2+17^2+18^2+19^2+20^2+21^2+22^2+23^2+24^2+25^2}{11}}$

$v_{rms}=20.24\ m/s$

Hence, this is the required solution.

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<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

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Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

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