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3 years ago

(I will give brainliest to the person who answered my question correctly!)

Chemistry

1 answer:

Elina [12.6K]3 years ago

7 0

Answer:yes

Explanation:

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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.

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

Molarity of 500 mL solution:

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

10 mL of this solution is diluted to 250 ml

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

Convert μM :

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

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

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

Molarity of 100 mL solution:

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

10 mL of this solution is diluted to 1000 ml

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

Convert μM :

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

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

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

Molarity of 100 mL solution:

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

5 mL of this solution is diluted to 1000 ml

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

Convert μM :

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

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

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

Molarity of 250 mL solution:

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

10 mL of this solution is diluted to 500 ml

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

Convert μM :

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

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

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

Molarity of 250 mL solution:

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

10 mL of this solution is diluted to 1000 ml

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

Convert μM :

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

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

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Answer:

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Explanation:

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potassium, k, has 19 protons, 20 neutrons, and 19 electrons. what is the charge of the atom’s nucleus?

mixer [17]

from the 19 protons

Answer:

+19

Explanation:

potassium, k, has 19 protons, 20 neutrons, and 19 electrons. what is the charge of the atom’s nucleus?

since only neutrons and protons are in the nucleus, the overall charge in the nucleus is +19 from the 19 protons

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