Answer:
1 litre is equal to 1000 millilitres
Covalent bonds can be classified as nonpolar and polar covalent given the electronegativity difference between two atoms (ΔEN).
Nonpolar covalent bond electrons are shared equally between two atoms, polar covalent bond electrons are shared unequally, atoms have partial charges, ionic bond electrons are completely transferred to one atom, full charges present. Therefore, the greater the electronegativity difference, the greater the bond polarity. Let's determine the types of bonds present in the compounds and arrange the ones with polar covalent in order of increasing ΔEN. Sulfur and oxygen are both nonmetals so the substance is covalent. Sulfur has EN = 2.5 and oxygen has EN = 3.5. Since there is an electronegativity difference, the S−O bonds in the substance can be classified as polar covalent bonds.
Learn more about polar covalent bond here:
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According to the balanced chemical equation:
4 HPO₃ + 12 C → 2 H₂ + 12 CO + P₄
4 moles of HPO₃ ---gives---> 12 moles of CO
2.73 moles of HPO₃ ---gives---> ? moles of CO
so number of moles of CO =
= 8.19 moles of CO
Number of molecules of CO = number of moles * Avogadro's number
= 8.19 * (6.022 * 10²³) = 4.93 * 10²⁴ molecules
<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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<span>The answer is Cs, no. 2.
Cesium (Cs) is an element which has an outermost electron with most energy in
the ground state.</span>
Cesium (Cs), Lithium (Li), Potassium
(K), and Sodium (Na) are elements which belong to the group 1 family, the
alkali metals. Each has a valance of 1 and wants to release/ lose this e-
(electron) to be isoelectronic with the nearest noble gas. Based on the
periodic trends, an atom’s radii will raise going down a group.
