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4 years ago

A 10 gram iron nail absorbed 125J of heat. What was the final temperature of the nail if the original temperature was 25°C?

Chemistry

1 answer:

valentina_108 [34]4 years ago

7 0

Answer:

52.1 degrees C

Explanation:

We need to use the equation: q = mCΔT, where m is the mass in grams, C is the specific heat capacity, and ΔT is the change in temperature.

Here, m = 10 g and q = 125 J. The heat capacity of iron is about 0.461 J/(g * C). And, our initial temperature is 25. So:

125 J = (10 g) * (0.461 J/(g * C)) * (T_f - 25)

Solving for T_f (final temp), we get: 52.1 degrees C

Hope this helps!

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Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but

alukav5142 [94]

Answer:

a) For nicotine, the protonated form is the present in stomach.

b) For caffeine, the neutral base form is the present in stomach.

c) For Strychinene, the protonated form is the present in stomach.

d) For quinine, the protonated form is the present in stomach.

Explanation:

In a basic dissociation for molecules with basic nitrogen, the equilibrium is:

A + H₂O ⇄ AH⁺ + OH⁻

<em>Where A is neutral base and AH⁺ is protonated form</em>

The basic dissociation constant, kb, is:

$K_{b} = \frac{[AH^+][OH^-]}{[A]}$

As pH in stomach is 2,5:

[OH] = $10^{-[14-pH]}$

[OH] = 3,16x10⁻¹² M

Thus:

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$ <em>(1)</em>

Using (1) it is possible to know if you have the neutral base or the protonated form, thus:

(a) nicotine Kb = 7x10^-7

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{7x10^{-7}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$221359 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For nicotine, the protonated form is the present in stomach

(b) caffeine,Kb= 4x10^-14

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{4x10^{-14}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$0,0127 = \frac{[AH^+]}{[A]}$

[AH⁺}<<<<[A]

For caffeine, the neutral base form is the present in stomach

(c) strychnine Kb= 1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$314456 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For Strychinene, the protonated form is the present in stomach

(d) quinine, Kb= 1.1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1,1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$347851= \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For quinine, the protonated form is the present in stomach.

I hope it helps!

7 0

3 years ago

What is the correct name

Inessa05 [86]

Answer:

The Answer is 2-methylpentane

Or the third one down.

Explanation:

5 0

3 years ago

Read 2 more answers

Given the reaction:

sineoko [7]

Answer:

B = 6 grams

Explanation:

please make me brainlist answer

4 0

3 years ago

Which statement is incorrect?

adoni [48]

I think it is d because it talks about other things the other ones finest talk about so go with d or c I hope you get it right good luck

8 0

3 years ago

Read 2 more answers

The P:OP:O ratio is the amount of inorganic phosphate incorporated into ATP per atom of oxygen consumed. It represents the coupl

Ray Of Light [21]

Answer:

For NADH; P:O = 2.5

For FADH ₂; P : O = 1.5

Explanation:

The P:O (phosphate:oxygen) ratio represents the amount of inorganic phosphate, Pi used per atom of oxygen consume to synthesize ATP.

The Chemiosmotic theory predicts H⁺:O and H⁺:ATP ratios. Experimentally these appear to be 10 and 4 respectively when NADH is the substrate, equivalent to a P:O ratio of 2.5, and 6 and 4 respectively for FAD-linked substrates (e.g. succinate), equivalent to a P:O ratio of 1.5.

1. Electron flow from NADH to O₂ pumps protons at three sites to yield 3 ATP (P:O = 2.5)

For NADH: 10 H ⁺ translocated/O (2e -)

ATP/2e - = (10 H⁺/ 4 H +) = 2.5

2. Succinate (via FADH2) bypasses site 1 giving 2 ATP (P : O = 1.5)

For FADH ₂= 6 H ⁺/O(2e - )

ATP/2e - = (6 H +/ 4 H +) = 1.5

7 0

3 years ago