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3 years ago

Plz Help! If 0.0025 mol NaNO3 forms during the reaction, what is the concentration of NaNO3 in the final solution?

Chemistry

1 answer:

OLEGan [10]3 years ago

3 0

Answer:

Hi! I hope this will help

you with your question, whether this is an assignment or not. The answer that I

get is 0.033 concentration.

If 0.0025 mol NaNo3 forms

during the reaction, the concentration of NaNO3 in the final solution should be

0.033.

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The sum of the first 10 terms of an arithmetic progression is 120 and the sum of first twenty is 840. find sum of first 30 terms

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Answer:

The sum of first 30 terms of the arithmetic progression is <u>2160.</u>

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For an arithmetic progression, the sum of first $n$ terms with first term as $a$ and common difference $d$ is given as:

$S_n=\frac{n}{2}(2a+(n-1)d)$

Now, it is given that:

$For\ n=10,S_n=120\\For\ n=20,S_n=840$

Now, plug in these values and frame two equations in $a\ and\ d$

$S_{10}=\frac{10}{2}(2a+(10-1)d)\\120=5(2a+9d)\\2a+9d=\frac{120}{5}\\2a+9d=24------------1$

$S_{20}=\frac{20}{2}(2a+(20-1)d)\\840=10(2a+19d)\\2a+19d=\frac{840}{10}\\2a+19d=84-----------2$

Now, we solve equations (1) and (2) for $a\ and\ d$ . Subtract equation (1) from equation (2). This gives,

$2a+19d-2a-9d=84-24\\19d-9d=60\\10d=60\\d=\frac{60}{10}=6$

Now, plug in the value of $d=6$ in equation (1) and solve for $a$ .

$2a+9(6)=24\\2a+54=24\\2a=24-54\\2a=-30\\a=\frac{-30}{2}=-15$

Plug in the values of $a=-15,\ n=30\ and\ d=6$ in the sum formula to find the sum of first 30 terms.

Now, the sum of first 30 terms is given as:

$S_{30}=\frac{30}{2}(2(-15)+(30-1)(6))\\S_{30}=15(-30+29(6))\\S_{30}=15(-30+174)\\S_{30}=15(144)=2160$

Therefore, the sum of first 30 terms of the arithmetic progression is 2160.

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3 years ago