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4 years ago

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In two or more complete sentences, describe how one of the occupations that you studied would use descriptive statistics when pe

rforming their job. Justify your answers in two or more complete sentences. in science

1 answer:

Korvikt [17]4 years ago

4 0

Answer:

I need an answer someone HELP MEEEEEEE

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The titration of 25.00 ml a 0.125 m hclo4 solution requires 27.07 ml of koh to reach the endpoint. what is the concentration of

agasfer [191]

Answer : The concentration of the $KOH$ is, 0.115 M

Explanation :

Using dilution law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of an acid = 1

$n_2$ = acidity of a base = 1

$M_1$ = concentration of $HClO_4$ = 0.125 M

$M_2$ = concentration of KOH = ?

$V_1$ = volume of $HClO_4$ = 25 ml

$V_2$ = volume of NaOH = 27.07 ml

Now put all the given values in the above law, we get the concentration of the $KOH$ .

$1\times 0.125M\times 25ml=1\times M_2\times 27.07ml$

$M_2=0.115M$

Therefore, the concentration of the $KOH$ is, 0.115 M

5 0

4 years ago

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

Neporo4naja [7]

Answer:

1) The dilution scheme will result in a 200μM solution.

2) The dilution scheme will not result in a 200μM solution.

3) The dilution scheme will not result in a 200μM solution.

4) The dilution scheme will result in a 200μM solution.

5) The dilution scheme will result in a 200μM solution.

Explanation:

Convert the given original molarity to molar as follows.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Consider the following serial dilutions.

1)

Dilute 5.00 mL of the stock solution upto 500 mL . Then dilute 10.00 mL of the resulting solution upto 250.0 mL.

<u>Molarity of 500 mL solution:</u>

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

<u>10 mL of this solution is diluted to 250 ml</u>

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

<u>Convert μM</u> :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

2)

Dilute 5.00 mL of the stock solution upto 100 mL . Then dilute 10.00 mL of the resulting solution upto 1000 mL.

<u>Molarity of 100 mL solution:</u>

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

<u>10 mL of this solution is diluted to 1000 ml</u>

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

<u>Convert μM</u> :

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

3)

Dilute 10.00 mL of the stock solution upto 100 mL . Then dilute 5 mL of the resulting solution upto 100 mL.

<u>Molarity of 100 mL solution:</u>

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

<u>5 mL of this solution is diluted to 1000 ml</u>

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

<u>Convert μM</u> :

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

4)

Dilute 5 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 500 mL.

<u>Molarity of 250 mL solution:</u>

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

<u>10 mL of this solution is diluted to 500 ml</u>

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

<u>Convert μM</u> :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

5)

Dilute 10 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 1000 mL.

<u>Molarity of 250 mL solution:</u>

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

<u>10 mL of this solution is diluted to 1000 ml</u>

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

<u>Convert μM</u> :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

7 0

3 years ago