Explanation:
According to the Henderson-Hasselbalch equation, the relation between pH and 
is as follows.
pH = 
where, pH = 7.4 and = 7.21
As here, we can use the nearest to the desired pH.
So, 7.4 = 7.21 + 
0.19 =
= 1.55
1 mM phosphate buffer means ![[HPO_{4}]](https://tex.z-dn.net/?f=%5BHPO_%7B4%7D%5D)
+ ![[H_{2}PO_{4}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DPO_%7B4%7D%5D)
= 1 mM
Therefore, the two equations will be as follows.
= 1.55 ............. (1)
+ = 1 mM ........... (2)
Now, putting the value of from equation (1) into equation (2) as follows.
1.55![[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DPO_%7B4%7D%5D%20%2B%20%5Btex%5D%5BH_%7B2%7DPO_%7B4%7D%5D)
= 1 mM
2.55 = 1 mM
= 0.392 mM
Putting the value of in equation (1) we get the following.
0.392 mM + = 1 mM
= (1 - 0.392) mM
= 0.608 mM
Thus, we can conclude that concentration of the acid must be 0.608 mM.
Answer:
I think no C or D
if wrong correct me plsssssss
Mark me brainliest plsss
Explanation:
December 21 marks the winter solstice in the Northern Hemisphere, but in 2020 the longest night of the year is also going to be a witness to an incredible astronomic event known as the “great conjunction”. ... This rare double planet sighting–or “Great Conjunction”–can be viewed from anywhere around the globe
Answer:
Acids change the color of litmus from blue to red.
They convert the color of Methyl Orange from Orange/Yellow to Pink.
Acids turn the pink color of Phenolphthalein to colourless.
Acids can conduct electricity.
Some Acids are highly corrosive in nature which means that they corrode or rust metals.
Explanation:
hope this helped
ur boi johnny joestar.
<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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