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horrorfan [7]
3 years ago
6

Which of the following statements concerning double covalent bonds is correct?

Chemistry
1 answer:
Alina [70]3 years ago
3 0

they involve tie sharing of 2 electrons pairs.

You might be interested in
Calculate the mass of 3.5 mol C6H6
Lilit [14]
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
4 0
3 years ago
A piece of high-density Styrofoam measuring 24.0 cm by 36.0 cm by 5.0 cm floats when placed in a tub of water. When a 1.5 kg boo
zheka24 [161]

Explanation:

The given data is as follows.

         Width of Styrofoam = 24.0 cm

          Length of Styrofoam = 36.0 cm

          Height of Styrofoam = 5.0 cm

Therefore, volume of the Styrofoam will be calculated as follows.

                  Volume = length × width × height

                                =  (36.0 × 24.0 × 5.0) cm^{3}

                                 = 4320 cm^{3}

or,                             = 4.32 \times 10^{3} cm^{3}

As Styrofoam partially sinks at 3.0 cm and total height of Styrofoam is 5.0 cm. Hence, height of Styrofoam above the water is (5.0 - 3 cm) = 2 cm.

So, volume of water displaced is as follows.

          24.0 cm × 36.0 cm × 2.0 cm

         = 1.73 \times 10^{3} cm^{3}

Hence, mass of displaced water is as follows.

                 mass = density × volume

                           = 1.00 g/cm^{3} \times 1.73 \times 10^{3} cm^{3}

                           = 1.73 \times 10^{3} g

Since, book is placed on the Styrofoam. Therefore, mass of water displaced is also equal to the following.

             Mass of water displaced = mass of book + mass of Styrofoam

                  1.73 \times 10^{3} g = 1500 g + mass of Styrofoam

                   (1730 - 1500) g = mass of Styrofoam

                   mass of Styrofoam = 230 g

Therefore, calculate the density of Styrofoam as follows.

                   Density = \frac{mass}{volume}  

                                 = \frac{230}{4.32 \times 10^{3} cm^{3}}

                                 = 53.24 \times 10^{-3} g cm^{-3}

Thus, we can conclude that the density of Styrofoam is 53.24 \times 10^{-3} g cm^{-3}.

4 0
3 years ago
Stuck on question 22
Klio2033 [76]
The answer is c. because you have to increase concentration of h2
7 0
3 years ago
One question please help!
Agata [3.3K]
<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>
3 0
3 years ago
Use the chart to determine which type of bond is formed between potassium (K) and chlorine (Cl).
iris [78.8K]
Potassium and Chloride forms an ionic bond.
(K+) + (Cl-) = KCl

Potassium is under Group IA (Alkali Metal), wherein elements under this group can easily lose electrons.

Chlorine is under Group VII (Halogens), in which these elements can gain electrons easily.

The inner shell electrons on potassium will merge with the outer shell of electrons of chlorine to make potassium chloride.
6 0
3 years ago
Read 2 more answers
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