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3 years ago

I am a metal that has 3 valence electrons and 123 neutrons

Chemistry

1 answer:

zlopas [31]3 years ago

3 0

Answer:

The answer is Boron

Explanation:

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True or false: the smaller the difference between the average experimental value and the correct (true) value the greater the ac

mel-nik [20]

Answer:

True

Explanation:

The accuracy level is usually determined by the difference between the experimental and correct value. It is important to note that the smaller the difference between the average experimental value and the correct (true) value, the more accurate it is.

When the difference is large then it means the accuracy level is low and not up to the required standard.

5 0

3 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.

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.

7 0

3 years ago

How much does calcite coast

Nikitich [7]

Answer:

It depends on what kind and how much. Some are about $100, and others are $1,000.

5 0

4 years ago

Waves that require a medium through which to travel are electromagnetic waves. True or False?

aev [14]

Answer:

True

Explanation:

Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate.

8 0

3 years ago

Read 2 more answers

What is the forward rate with continuous compounding for a three-month period starting in one year?

Tamiku [17]

Answer:8.7%

Explanation:

5 0

2 years ago