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e-lub [12.9K]
3 years ago
15

A neutral atom possesses an atomic number of 15 and an atomic mass of 31. Three electrons are gained. What is the result of this

conversion?
A. a positively charged ion

B. a new isomer of the element

C. a new isotope of the element

D. a negatively charged ion
Chemistry
2 answers:
Nonamiya [84]3 years ago
6 0

Answer:

D. a negatively charged ion

Explanation:

In the neutral state of any element the number of electrons and number of protons are equal in the number. Here in this case since number of electrons are 15 ( equal to atomic number) therefore number of protons are also 15. When 3 electrons are gained by the element this results in a  a negatively charged ion. The ion will have -3 charge over it. It either means excess of three electrons or deficiency of three protons.

12345 [234]3 years ago
5 0
On a neutral state, the number of proton of a substance will be equal to the number of its electron (in this case, it has 15 electrons). When three electrons are gained, the total number of the electron will change to 18.

Therefore this result in the substance becomes - D. A negatively charged ion of 3-

Hope this helps!
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Enthalpy of combustion of ethyl alcohol, C2 H5 OH, is – 950 kJ mol–1. How much heat is evolved when one gram of ethyl alcohol bu
melomori [17]
<h3>Answer:</h3>

20.62 Kilo-joules

<h3>Explanation:</h3>
  • The Enthalpy of combustion of ethyl alcohol is -950 kJ/mol.
  • This means that 1 mole of ethyl alcohol evolves a quantity of heat of 950 Joules when burned.

Molar mass of ethyl ethanol = 46.08 g/mol

Therefore;

46.08 g of  C₂H₅OH evolves heat equivalent to 950 kilojoules

We can calculate the amount of heat evolved by 1 g of C₂H₅OH

Heat evolved by 1 g of C₂H₅OH  = Molar enthalpy of combustion ÷ Molar mass

                                      = 950 kJ/mol ÷ 46.08 g/mol

                                      = 20.62 Kj/g

Therefore, a gram of C₂H₅OH  will evolve 20.62 kilo-joules of heat

7 0
2 years ago
Let’s say you have 3 UV active spots in the crude material and co-spot TLC plate. One has the same Rf value as the starting mate
Troyanec [42]

Answer:

Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)  

Explanation:

Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.

In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.

8 0
2 years ago
Read 2 more answers
Which molecule has 6 carbon atoms
densk [106]
Glucose molecule has 6 carbon atoms
8 0
3 years ago
The usual units of density are : <br> cm3/ g<br> cm2/g<br> g/cm<br> Nm
Ad libitum [116K]
The usual units of density are g/cm.

You may have also seen g/mL used for density. Keep in mind that 1 cm = 1 mL.
3 0
3 years ago
Read 2 more answers
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
2 years ago
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