問上面照片的題目，何者不是構成蛋白質的主要成分原素？

Be sure to answer all parts. what are the concentrations of hso4−, so42−, and h in a 0. 31 m khso4 solution? (hint: h2so4 is a s

disa [49]

At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R $HSO_4^-$ ⇄ $H^+ + SO_4^2^-$

I $0.14$

C $- x$ $+x$ $+x$

E $0.14-x$ $x$ $x$

$K_a = 1.3$ × $10^-^2$ for $HSO^-_4$ . As a result,

$\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a$

$K_a$ is large. It is no longer valid to approximate that $[HSO^-_4]$ at equilibrium is the same as its initial value.

$\frac{x^2}{0.14-y} = 1.3 * 10^-^2$

$x^2+1.3*10^-^2x - 0.14$ × $1.3$ × $10^-^2= 0$

Solving the quadratic equation for $x , x \geq 0$ since $x$ represents a concentration;

$x=0.0366538$

Then, round the results to 2 significant figure;

$[SO_4^2^-] = x = 0.037 mol. L ^-^1$
$[H^+] = x = 0.037 mol. L ^-^1$
$[HSO_4^-] = 0.14 - x = 0.10 mol. L ^-^1$

Learn more about concentration here:

brainly.com/question/14469428

#SPJ4

3 0

1 year ago

Which type of radioactive decay can pass through the body?

Mkey [24]

Q. Which type of radioactive decay can pass through the body?

A. Gamma Rays

3 0

3 years ago

Description of how the combined gas law must be modified to introduce the number of moles

telo118 [61]

So the ideal gas law is pv=nrt
The letter n stands for the number of moles. divide both side by rt to isolate.
pv/rt=n

8 0

3 years ago

Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate gives, after protonolysis, a good yield of a mixture

tekilochka [14]

Answer:

use google and use the first link

Explanation:

6 0

3 years ago

C5H12 + 8O2 → 5CO2 + 6H2O<br><br> Calculate the energy change for the reaction:

KATRIN_1 [288]

C5H12 (l) + 8O2 (g) ----> 5CO2 (g) + 6H2O (l)
Delta H = -3505.8 kJ/mol

C (s) + O2 (g) -----> CO2 (g)
Delta H = -393.5 kJ/mol

H2 (g) + (1/2)O2 (g) ------> H2O (l)
Delta H = -286 kJ/mol

Possible answers:
a. +35 kJ/mol
b. + 1,073 kJ/mol
c. -4,185 kJ/mol
d. -2,826 kJ/mol
e. -178 kJ/mol

8 0

3 years ago