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3 years ago

12

. Explain why some desert animals excrete uric acid rather than ammonia.(2 marks)

1 answer:

avanturin [10]3 years ago

5 0

Answer:

AFAIK

Explanation:

uric acid is much less toxic than ammonia, hence bigger concentrations of it are tolerated in the body. This means you can excrete it while excreting very little water - beneficial wherever water's not abundant.

There's a tradeoff though, uric acid requires more energy to synthesize than ammonia, so pretty much all fish, say, excrete ammonia rather than uric acid - it's no problem to dilute ammonia since there's no water shortage.

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At constant force, acceleration varies (directly, inversely) with mass. When subjected to the same amount of net external force,

lora16 [44]

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3 years ago

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When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

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3 years ago

Sidana [21]

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3 years ago

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PLZ ANSWER FIRST ONE TO ANSWER GETS BRAINLIEST!!!!!!

Roman55 [17]

The answer is “Heat Conduction”

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3 years ago

Methanol has a normal boiling point of 64.6C and a heat of vaporization of 35.2 kJ/mol. What is the vapor pressure (in Torr) of

DENIUS [597]

Answer:

vapor pressure of methanol at 12.0C = 75.09 torr

Explanation:

Using Clausius Clapeyron equation

, we have that

ln (P2/P1)= (ΔHvap /R) (1/T1 - 1/T2)

Given

At Normal boiling point,

Temperature T1= 64.6°C = 64.6 + 273 = 337.6 K, Pressure,P1 = 1 atm

Heat of vaporization = 35.2 kJ/mol

Changing to J/mol

=35.2 x 1000= 35200 J/mol

Temperature , T2 = 12.0oC = 12 + 273 = 285 K

Using gas constant, R = 8.314 J/mol.K

ln (P2/P1)= -(ΔHvap /R) (1/T1 - 1/T2)

ln (P2/ 1 atm) = (35200 J/mol/ (8.314 J/mol.K) X( 1/337.6 - 1/285)

ln (P2/ 1 atm) =4,233.822 X (0.00296-0.003508)

ln (P2/ 1 atm) = 4,233.822468 x-0.0005466866

ln (P2/ 1 atm)= -2.31457

P2 = e^⁻2.31457 x 1 atm

P2=0.098808atm

= 0.098808atm x760 = 75.09 torr

7 0

3 years ago