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4 years ago

How many mL of 6.00M HCl are needed to neutralize 50.00mL of 3.25M NaOH?

Mathematics

1 answer:

MakcuM [25]4 years ago

5 0

Answer:

0.027L or 27.1mL

Step-by-step explanation:

v1m1=v2m2

v1x6M=.05Lx3.25M

V1=0.027L or 27.1mL

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Take the derivative to get

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Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

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Take the derivative one more time:

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$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

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