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yanalaym [24]
3 years ago
7

II . Explain the differences among these general processes of respiration.

Chemistry
1 answer:
MAVERICK [17]3 years ago
5 0
B is the answer to this q. Cellular
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The six pollutants regulated under the Clear Air Act _____ between 1970 and 2001, while energy consumption increased 42 percent.
pochemuha

Answer:

increased only 6 percent

Explanation:

The six pollutants regulated under the Clear Air Act increased only 6 percent between 1970 and 2001, while energy consumption increased 42 percent.

7 0
3 years ago
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Chemistry
madam [21]

Answer: 1. 3.914 × ^10-4 | 2. 4.781 × ^10-1

Explanation:

7 0
3 years ago
Before the fire, the forest consists of large trees. After the fire, there is only ash. Explain what the law of conservation of
Tasya [4]

Answer: explained below

Explanation:

Matter can change form through physical and chemical changes, but through any of these changes, matter is conserved. The same amount of matter exists before and after the change—none is created or destroyed.

5 0
3 years ago
A solution has [H+] = 2.35 × 10-3 M. Find the [OH-] for this solution
Aleks04 [339]

Answer:

[OH-] for this solution is 4.255*10^-12

Explanation:

We are given

[H+] = 2.35 × 10-3 M

we need to find the concentration of [OH-]

we know from Equilibrium

[H+][OH-] = 10^-14

[OH-] = 10^14/2.35*10^10^-3

[OH-] = 0.4255*10^-11

[OH] = 4.255*10^-12

Therefore the Concentration  of [OH-] for this solution is 4.255*10^-12

3 0
4 years ago
<img src="https://tex.z-dn.net/?f=H_2PO_4%5E-%28aq%29%20%5Crightarrow%20H%5E%2B%28aq%29%20%2B%20HPO_4%5E%7B2-%7D%28aq%29" id="Te
klasskru [66]

Answer:

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

pH = 7.21 + 0.84

pH = 8.05

4 0
3 years ago
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