Answer:
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
Explanation:
The equation is given as;
2HCl(aq) + Zn(s) + H2(g) + ZnCl2(aq)
In writing an ionic equation, only the aqueous compounds dissociates into ions. This means HCl and ZnCl2 would dissociate to form ions.
This is given as;
2H+ + 2Cl- + Zn(s) --> H2(g) + Zn2+ + 2Cl-
The correct option is;
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
The answer is flourine
flourine some what sounds like flow so
flourine is a flowing element
<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 :)
</span>
Answer:
lattice parameter = 5.3355x10^-8 cm
atomic radius = 2.3103x10^-8 cm
Explanation:
known data:
p=0.855 g/cm^3
atomic mass = 39.09 g/mol
atoms/cell = 2 atoms
Avogadro number = 6.02x10^23 atom/mol
a) the lattice parameter:
Since potassium has a cubic structure, its volume is equal to:
v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]
substituting values:
v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3
but as the cell volume is
a^3 =v
![a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7Bv%7D%3D%5Csqrt%5B3%5D%7B1.5189x10%5E%7B-22%7D%20%7D%20%3D%205.3355x10%5E-8)
cm
for a BCC structure, the atomic radius is equal to
