Answer:
6.23 x 10^23 molecules
Explanation:
First find the number of moles of BH3 from the information given. We know the amount of grams present and we can find the molar mass which is 13.84.
We know that moles is grams divided by molar mass so we get 14.32/13.84 which is 1.03 moles.
Finally, to figure out the number of molecules, we multiply 1.03 by Avogadro's number which is 6.022x10^23 and we get 6.23x10^23 molecules.
The mass of water in the tank, given the data from the question is 549594 g
<h3> Description of mole </h3>
The mole of a substance is related to it's mass and molar mass according to the following equation:
Mole = mass / molar mass
<h3>How to determine the mass of water in the tank</h3>
From the question given above, the following data were obtained:
- Mole of water = 30533 moles
- Molar mass of water = 18 g/mol
- Mass of water = ?
The mass of the water can be obtained as follow:
Mass = mole × molar mass
Mass of water = 30533 × 18
Mass of water = 549594 g
Learn more about mole:
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Answer:
Ka = ( [H₃O⁺] . [F⁻] ) / [HF]
Explanation:
HF is a weak acid which in water, keeps this equilibrium
HF (aq) + H₂O (l) ⇄ H₃O⁺ (aq) + F⁻ (aq) Ka
2H₂O (l) ⇄ H₃O⁺ (l) + OH⁻ (aq) Kw
HF is the weak acid
F⁻ is the conjugate stron base
Let's make the expression for K
K = ( [H₃O⁺] . [F⁻] ) / [HF] . [H₂O]
K . [H₂O] = ( [H₃O⁺] . [F⁻] ) / [HF]
K . [H₂O] = Ka
Ka, the acid dissociation constant, includes Kwater.
Answer:
The pond has more energy because I is so much larger that the Cup of boiling water. Since that mass of the pond is so much larger, It is generating more energy than a boiling cup of water.
<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.
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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).
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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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