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Licemer1 [7]
3 years ago
8

How does ph affect photosynthesis?

Chemistry
1 answer:
liraira [26]3 years ago
4 0
Photosynthesis is the process by which plants create their food. It involves the operation of enzymes in the plant cells , and they work best at certainpH levels. ... Thus, as the plant's pH drifts away from the best pH, the rate of photosynthesis will decrease.
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Is vegetable soup a substance or a mixture
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Vegetable soup is a mixture. 
6 0
2 years ago
Read 2 more answers
calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate
kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

brainly.com/question/9240031

#SPJ4

4 0
1 year ago
A. Identify the structure drawn below.
nadya68 [22]

Answer:

Hexane

Explanation:

You have a carbon structure with only single bonds.  This means that the name will end in -ane.

There are 6 carbon atoms.  This means that the name will begin with hex-.

The structure is hexane.

7 0
3 years ago
How do scientists design a system?
Shtirlitz [24]

Answer: D

Explanation:

3 0
3 years ago
A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of t
kobusy [5.1K]

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point <u>F</u>)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

5 0
3 years ago
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