Answer:
Group B
Explanation:
The control group of an experiment is considered to be the "normal" because it does not receive the expeiemental treatment and hence is used to compare with the experimental group. The control and experimental groups are similar in every other aspect with the exception of the "INDEPENDENT VARIABLE". The independent variable is not changed in the control group.
In this experiment, the control group is GROUP B, which was given a plain gatorade without adding sugar. Sugar is the independent variable here, and it was not included in GROUP B, meaning that GROUP B is the CONTROL GROUP. On the other hand, the experimental group is the GROUP A.
I did the math i got 220 grams
The pH of an aqueous solution with a hydrogen ion concentration of [H+]=1. 2×10^−8 m is 7.92.
The entire form of pH is potential of Hydrogen. pH is called the poor logarithm of H+ ion attention. hence the that means of the name pH is explained because the energy of hydrogen. pH describes the concentration of the hydrogen ions in a solution and it's miles the indicator of acidity or basicity.
pH is an important amount that reflects the chemical conditions of an answer. The pH can manage the provision of vitamins, organic features, microbial activity, and the conduct of chemical substances.
pH stands for capacity hydrogen and it's used to explain the acidity or alkalinity of a substance. The pH scale is measured from 1 to fourteen. 7 is impartial and the lower the quantity, the extra acidic and the higher, the greater alkaline.
explaination,
[H+]=1. 2×10^−8 m
log[h+] = 1. 2×10^−8 m
= 8 - log (1.2)
= 8 - 0.079
PH = 7.921
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A chemical element bonded to an identical chemical element is not a chemical compound since only one element, not two different elements, is involved. Examples are the diatomic molecule hydrogen (H2) and the polyatomic molecule sulfur (S8).
Henderson–Hasselbalch equation is given as,
pH = pKa + log [A⁻] / [HA]
-------- (1)
Solution:
Convert Ka into pKa,
pKa = -log Ka
pKa = -log 1.37 × 10⁻⁴
pKa = 3.863
Putting value of pKa and pH in eq.1,
4.29 = 3.863 + log [lactate] / [lactic acid]
Or,
log [lactate] / [lactic acid] = 4.29 - 3.863
log [lactate] / [lactic acid] = 0.427
Taking Anti log,
[lactate] / [lactic acid]
= 2.673
Result:
2.673 M
lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.
