Answer:
The volume in the first tank = 0.32 
The volume in the second tank = 2.066
The final pressure of the mixture = 203.64 K pa
Explanation:
<u>First Tank </u>
Mass = 2 kg
Pressure = 550 k pa
Temperature = 25 °c = 298 K
Gas constant for nitrogen = 0.297 
From the ideal gas equation
P V = m R T
550 × V = 2 × 0.297 × 298
V = 0.32
This is the volume in the first tank.
<u>Second tank</u>
Mass = 4 kg
Pressure = 150 K pa
Temperature = 25 °c = 298 K
Gas constant for oxygen = 0.26
From the ideal gas equation
P V = m R T
150 × V = 4 × 0.26 × 298
V = 2.066
This is the volume in the second tank.
This is the iso thermal mixing. i.e.
----- (1)
Put this value in equation (1)
× 2.386 = 550 × 0.32 + 150 × 2.066
= 203.64 K pa
Therefore the final pressure of the mixture = 203.64 K pa
The answer is D I hope this helps you !
Elements with three p-electrons....
That would be N, P, As, Sb, and Bi -- elements in group 15
For example, energy diagram showing "empty" orbitals up through the 3p.
.....3p __ __ __
3s __
.....2p __ __ __
2s __
1s __
Energy diagram of phosphorous showing three unpaired electrons in 3p-sublevel
.....3p ↑_ ↑_ ↑_
3s ↑↓
.....2p ↑↓ ↑↓ ↑↓
2s ↑↓
1s ↑↓
According to Hund's rule, the electrons singly occupy the p-orbitals, and all have the same spin.
Answer:
Less than one gram
Explanation:
Since there is no whole number before the decimal it means that the number is less than whole meaning it is less than one gram
<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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