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Karolina [17]
3 years ago
14

Clathrate hydrate research is done by Professor Kend Janda at UC Irvine. Look up clathrate hydrate with a general internet searc

h and in Odyssey (Applied Chemistry, #86 Clathrates). The ethyl (CH3CH2 –) end of the ethanol molecule (CH3CH2OH) is behaving much like the propane (CH3CH2CH3) and methane (CH4) gases that Prof. Janda\'s group uses to form clathrates. These hydrocarbons (propane, methane, and ethyl) are all nonpolar and, therefore, do not contain any significant dipoles. The _________________ creates a cage where the water is the _________________ for/of the hydrocarbon.
Chemistry
1 answer:
lidiya [134]3 years ago
3 0

Answer:

water, host

Explanation:

Clathrate hydrates are the crystalline water-based solids which physically resembles with the ice, in which the small non-polar molecules or smaller polar molecules with the large hydrophobic moieties are usually trapped inside the "cages" of the hydrogen bonded as well as frozen water molecules.  

<u>Clathrate hydrates are the clathrate compounds in which water creates the cage and the water behaves as a host molecule and guest molecule is typically a gas or liquid.</u>

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What would happen if this cell were moved from pure water to a seawater solution?
Reika [66]
<span>The cell would swell and burst, because the 20% salt solution is hypotonic with respect to the cell, causing a net movement of water into the cell. Example, water moves from the blood filtrate that will form urine. The same will do while pure water moved to seawater.</span>
3 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
What is the density (g/L) of NH3 at 105 kpa and 25 C ( 0.721 g/mol)?
jenyasd209 [6]

Answer:

0.721 g/L

Explanation:

Ideal gas equation ->PV= nRT   ; n= mass (m)/ molar mass (M);

densitiy= mass (m)/ volume (V)

PV= (m/M)*RT  -> PVM= mRT   -> PM/RT= m/V   -> PM/RT=d

We need to put in SI units

105 Kpa= 1.04 atm

25°C= 298 K

d= (1.04 atm * 17 g/ mol)/(0.0821 * 298 K) = 0.721 g/L

3 0
3 years ago
Which BEST explains why the sun appears reddish at sunrise? A) Blue wavelengths of light have already been scattered before reac
romanna [79]
<span>The answer is A. This is because sunlight is composed of various wavelengths in the electromagnetic spectrum. In the visible light spectrum, red light has a higher frequency (lower wavelength) than blue light that has a lower frequency (higher wavelength). Therefore red light has higher energy than blue light hence able to penetrate more into the atmosphere before scattering, compared to blue light. At sunrise and sunset, due to the relative angle  of the sun to the observer, the sun rays have to penetrate a thicker layer to the atmosphere before reaching the observer</span>




7 0
3 years ago
100 grams of iron ore contains 30.06% oxygen find the empirical formula​
Serga [27]

Answer: Fe2O3

Explanation:

100 grams of iron ore = 30.06% oxygen

100 - 30.06 = 69.94% iron

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So it comes out to be Fe2O3

8 0
2 years ago
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