Answer:
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Answer:
0,727 mL
Explanation:
We add x mL of the 95% ethanol solution to 1 mL: C1=0,95 and V1=x mL
As we already had 1 ml of solution, now we'll have (1 + x) mL of the 40% ethanol solution: C2=0,4 and V2=1+X mL
Now, using the C1V1=C2V2 equation:
(0,95)x=0,4(x+1)
0,95x=0,4x + 0,4
0,95x-0,4x=0,4
0,55x=0,4
x=0,4/0,55=0,727 mL
Therefore, we need to add 0,727 mL of a 95% ethanol solution to the enzyme solution to obtain a final concentration of 40% ethanol.
Answer:
Explanation:
Hello,
In this case, the combustion of octane is chemically expressed by:
In such a way, due to the 25/2:8 molar ratio between oxygen and carbon dioxide, we can compute the yielded grams of carbon dioxide (molar mass 44 g/mol) as shown below:
Best regards.
Both chloro and methyl groups axial
More stable
Higher concentration
