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zloy xaker [14]
3 years ago
8

calculate the equilibrium concentration for the nonionized bases and all ions in a solution that is 0.25M

Chemistry
1 answer:
Anika [276]3 years ago
5 0

Answer:

The equilibrium concentration of [CH₃NH₂] = 0.23965 M.

The equilibrium concentration of CH₃NH₃⁺ and  OH⁻ =  0.01035 M respectively.

Explanation:

The first step is to write out the dissociation reaction. Therefore, the equation showing the dissociation is given as below;

CH₃NH₂ + H₂0   <--------------------------------------------------------> CH₃NH₃⁺ + OH⁻.

Kindly note that ''<----------->'' arrow shows that the reaction is an equilibrium reaction.

Therefore, at the start of the reaction [that is time, t =0], we have that the concentration of CH₃NH₂ = 0.25M, thus, the concentration of CH₃NH₃⁺  and  OH⁻ is zero respectively at this time, t =0.

At equilibrium, the concentration of CH₃NH₂ = 0.25M - x, thus, the concentration of CH₃NH₃⁺  and  OH⁻ is x respectively.

Therefore, kb =  4.47 × 10-4 = [CH₃NH₃⁺ ][OH⁻]/[CH₃NH₂].  Hence, slotting in the values into this equilibrium equation showing the relationship between kb and concentration of the species involved, we have that;

kb = 4.47 × 10⁻⁴ = x² /0.25 - x.

x² +  4.47 × 10⁻⁴x - 1.1175  × 10⁻⁴ = 0.

Solving this quadratic equation gives us the value of x as 0.01035 M.

Thus, the concentration of [CH₃NH₂] = 0.25 M - 0.01035 M = 0.23965 M

The equilibrium concentration of [CH₃NH₂] = 0.23965 M.

The equilibrium concentration of CH₃NH₃⁺ and  OH⁻ =  0.01035 M respectively.

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Answer: Option (A) is the correct answer.

Explanation:

Force acting on a dam is as follows.

                  F = \frac{1}{2}\rho g\omega H^{2} .......... (1)

Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.

                F' = \frac{1}{2}\rho g\omega (2)^{2}

               F' = \frac{1}{2}\rho g\omega 4 ........... (2)

Now, dividing equation (1) by equation (2) as follows.

          \frac{F}{F'} = \frac{\frac{1}{2}\rho g\omega H^{2}}{\frac{1}{2}\rho g\omega 4}  

Cancelling the common terms we get the following.

                 \frac{F}{F'} = \frac{1}{4}    

                   4F = F'

Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.

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What happens as a result of the photoelectric effect? A.light is produced
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One question please help!
Agata [3.3K]
<span>1 trial : you have nothing to compare the result with - you don't know if it's a mistake.
2 trials : you can compare results - if very different, one may have gone wrong, but which one?
3 trials : if 2 results are close and 3rd far away, 3rd probably unreliable and can be rejected.

******************************

First calculate the enthalpy of fusion. M, C and m,c = mass and specific heat of calorimeter and water; n, L = mass and heat of fusion of ice; T = temperature fall.

L = (mc+MC)T/n.

c=4.18 J/gK. I assume calorimeter was copper, so C=0.385 J/gK.

1. M = 409g, m = 45g. T = 22c, n = 14g
L = (45*4.18+409*0.385)*22/14 = 543.0 J/g.

2. M = 409g, m = 49g, T = 20c, n = 13g
L = (49*4.18+409*0.385)*20/13 = 557.4 J/g.

3. M = 409g, m = 54g, T = 20c, n = 14g
L = (54*4.18+409*0.385)*20/14 = 547.4 J/g.

(i) Estimate error in L from spread of 3 results.
Average L = 549.3 J/g.
average of squared differences (variance) = (6.236^2+8.095^2+1.859^2)/3 = 35.96
standard deviation = 5.9964
standard error = SD/(N-1) = 5.9964/2 = 3 J/g approx.

% error = 3/547 x 100% = 0.5%.

(ii) Estimate error in L from accuracy of measurements:
error in masses = +/-0.5g
error in T = +/-0.5c

For Trial 3
M = 409g, error = 0.5g
m = 463-409, error = sqrt(0.5^2+0.5^2) = 0.5*sqrt(2)
n =(516-463)-(448-409)=14, error = 0.5*sqrt(4) = 1.0g
K = (mc+MC)=383, error = sqrt[2*(0.5*4.18)^2+(0.5*0.385)^2] = 2.962

L = K*T/n
% errors are
K: 3/383 x 100% = 0.77
T: 0.5/20 x 100% = 2.5
n: 1.0/14 x 100% = 7.14

% errors in K and T are << error in n, so we can ignore them.
% error in L = same as in n = 7% x 547.4 = 40 (always round final error to 1 sig fig).

*************************************

The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.
Both are very far above accepted figure of 334 J/g, so there is at least one systematic error in the experiment or the calculations.
eg calorimeter may not be copper, so C is not 0.385 J/gK. (If it was polystyrene, which absorbs/ transmits little heat, the effective value of C would be very low, reducing L.)
Using +/- 40 is probably best (more cautious).
However, the spread in the actual results is much smaller; try to explain this discrepancy - eg
* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.
* measurements were more accurate than I assumed (eg masses to nearest 0.1g but rounded to 1g when written down).

Other sources of error:
L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?
* it is suspicious that all final temperatures were 0c - was this actually measured or just guessed? a higher final temp would reduce L.
* we have assumed initial and final temperature of ice was 0c, it may actually have been colder, so less ice would melt - this could explain small values of n
* some water might have been left in container when unmelted ice was weighed (eg clinging to ice) - again this could explain small n;
* poor insulation - heat gained from surroundings, melting more ice, increasing n - but this would reduce measured L below 334 J/g not increase it.
* calorimeter still cold from last trial when next one started, not given time to reach same temperature as water - this would reduce n.
Hope This Helps :)
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