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Kaylis [27]
3 years ago
12

How many liters of 2.0 M HCl are required to exactly neutralize 1.5 L of 5.0 M NAOH?

Chemistry
1 answer:
boyakko [2]3 years ago
7 0

Answer:

3.25 L

Explanation:

A 2.0M solution of HCl will have 2 moles of HCl for each liter of solution. 1.5 liters of 5.0M NaOH has a total of 5*1.5=7.5 moles of NaOH. Therefore, you would need 7.5/2=3.25 liters of HCl. Hope this helps!

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Yes because the protons are the same but the neutrons change
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You are a marine biologist who wants to study what type of zooplankton (microscopic animals that live in the ocean and feed larg
timama [110]

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The advantage of the compound light microscope over the dissecting microscope is the magnification power of the telescope. Compound light microscope magnifies from 40x up to 1,000x while dissecting microscope magnifies up to 40 x only. In this regard, more magnification power is advantageous to view smaller objects. 

4 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
The change in enthalpy that occurs in the process of converting reactants to products in a chemical reaction is called the _____
zheka24 [161]

Answer:

∆H or Enthalpy of the reaction

Explanation:

If ∆H is +ve

  • Reaction is exothermic
  • Example:-Combustion, mixing sodium/potassium in water

If ∆H is -ve

  • Reaction is endothermic
  • Ex:-Melting of ice
6 0
2 years ago
Read 2 more answers
Why is it important to test the crucible for cracks before heating it?
solmaris [256]

It important to test the crucible for cracks before heating it because it can break while performing the chemical reaction.

The crucible and the lid should be inspected to check for the before heating it to prevent the breaking while performing the reactions. if there are some breaks in the crucible then if we heat the crucible it can lead to breaking of the whole crucible and the material  from the crucible to come out and it will come in contact with the lid.

The crucible plays a important role while preforming the reactions, so. it important to check the cracks of the crucible before the heating of the crucible.

To learn more about crucible here

brainly.com/question/10237849

#SPJ4

7 0
1 year ago
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