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strojnjashka [21]
4 years ago
7

(PLZ HELP ME ON THIS!!!) Which of the following will change if you apply unbalanced forces to an object?

Chemistry
2 answers:
S_A_V [24]4 years ago
8 0
D) Its position changes because the unbalanced forces move the object.
vladimir1956 [14]4 years ago
5 0
It will change its position because it will have to move.
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HELP PLEASE IM GIVING BRAINLISTAs the amount of salt in water increases, the___________ of the water increases.
kari74 [83]
The answer is density
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3 years ago
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The number of ATPs per NADH generated by the electron transport system is higher than the number generated per FADH2 because ___
photoshop1234 [79]

Answer:

FADH2 has a lower (less negative) redox potential than NADH does

Explanation:

Flavin Adenine Dinucleotide (FAD) and Nicotinamide Adenine Dinucleotide (NAD) are redox cofactors that play important functions for mitochondrial activity and cellular redox balance. Both coenzymes exist in two forms: an oxidized and a reduced, which are abbreviated as NAD/FAD and NADH/FADH2, respectively. These reduced forms (NADH and FADH2) are produced in the Krebs cycle during respiration. FADH2 has lower redox potential than NADH because FADH2 is only capable of activating 2 proton pumps, while NADH can activate 3 proton pumps during the electron transport chain, thereby FADH2 generates a minor number of ATP molecules than NADH.

4 0
4 years ago
Determine molar mass of Mn (ClO3)3
goldenfox [79]
M(Mn(ClO3)3)=(54.938)+(35.45x3)+(15.999x9)
M(Mn(ClO3)3)=305.279 g/mol
8 0
3 years ago
A) If Kb for NX3 is 4.5×10^−6, what is the pOH of a 0.175 M aqueous solution of NX3 ?
sukhopar [10]

3.08 is the pOH of a 0.175 M aqueous solution of NX_3.

0.215% is the per cent ionization of a 0.325 M aqueous solution of NX_3

<h3>What is pH?</h3>

pH is a measure of how acidic/basic water is.

A)

NX_3 + H_2O →NHX_3^+ + OH^-

Kb = 4.5 x10^-6

Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).

concentration of (NH₄⁺) = concentration of (OH⁻) = x.

x² = Kb x concentration of (NH₃)

x² = 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)

= 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

=- log ( 8.367 × 10⁻⁴)

= 3.08

B)

Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.

Concentration of (NX₃) = 0.325 M.

Kb = 4.5 x 10⁻⁶.

[NX₃H⁺] = [OH⁻] = x.

[NX₃] = 0.325 M - x.

Kb = [NX₃H⁺] x [OH⁻] ÷  [NX₃].

4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).

x = 0.0007 M.

Per cent of ionization:

α = 0. 0007 M ÷ 0. 325 M x 100%

= 0.215%.

Hence,

3.08 is the pOH of a 0.175 M aqueous solution of NX_3.

0.215% is the per cent ionization of a 0.325 M aqueous solution of NX_3

Learn more about pH here:

brainly.com/question/12353627

#SPJ1

5 0
2 years ago
What volume of 12 M HCl solution is needed to make 2.5 L of 1.0 M HCI?
Jlenok [28]

Answer:

0.21 L of 12 M HCL

Explanation:

CV=CV

(12)(x)=(1.0)(2.5)

x=0.21

8 0
4 years ago
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