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ludmilkaskok [199]
2 years ago
6

Freeeeeeee pointsssssssssssssssss have a good day owo :) going once going twice andddddddddddddddddd

Biology
2 answers:
vagabundo [1.1K]2 years ago
3 0
Thank you for the points i appreciate it : )
mihalych1998 [28]2 years ago
3 0

Answer:

awww tysm i really needed this, we need more people like you<3

Explanation:

have a nice day and stay safe! 0w0

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Mars2501 [29]

Answer:

Adrenaline

Explanation:

<em>Adrenal</em>ine is released into the bloodstream by the <em>Adrenal Cortex</em> during stress.

7 0
2 years ago
Planet C has a tilt of zero degrees. What seasonal changes would be expected on this planet?
Contact [7]
The awnser is: No change in temperatures between seasons
7 0
3 years ago
2. List three sources of error that could account for the differences between your values for the enthalpy of fusion of water an
Dvinal [7]

1 trial :  nothing is given for result comparision - so we have no idea if it's a mistake.

2nd trial : The results can be compared - if varies, one may go wrong, but which one?

3rd trial : If 3rd result is different from 1st and 2nd, it is unreliable.

calculating 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. assuming copper calorimeter , 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.

squared differences average (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 ignore them.

% error in L = same as in n = 7% x 547.4 = 40

The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.

Both are very far above  334 J/g, so there is at least one systematic error  

e.g: 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 best.

However, the spread in the actual results is much smaller

* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.

<h3>Other sources of error: </h3>

L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?

* we have assumed initial and final temperature of ice was 0c, it may actually have been colder, so less ice would melt -which 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.

3 0
3 years ago
Which of the following is a common treatment for type III hypersensitivity reactions?
mestny [16]

Answer:

Anti-inflammatory steroid treatments

Explanation:

The exaggerated or hyper active response of the immune system causes hyper sensitive reactions in the body. These immune reactions are generally uncomfortable and harmful for the body.

Type III hypersensitivity reaction mainly occur due to the neutrophils, IgG and complement. This hyper sensitivity reaction can be treated by anti-inflammatory steroid treatments as the neutrophils are the main mediators of this reaction.

Thus, the correct answer is option (1).

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3 years ago
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LenaWriter [7]
Switch to anaerobic respiration
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3 years ago
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