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1 year ago

John is carefully pouring a chemical into a beaker when the beaker slips and breaks

1 answer:

5 0

Following laboratory safety protocols such as wearing personal protective equipment will protect John when the accident occurred.

<h3>What are laboratory safety protocols?</h3>

Laboratory safety protocols are the protocols put in place to ensure safety in the laboratory.

Laboratory safety protocols include the following:

always wear personal protective equipment in the laboratory
do not play in the laboratory
do not eat in the laboratory

Following laboratory safety protocols will help protect us from accidents which occur in the laboratory.

What happened when john was carefully pouring a chemical into a beaker when the beaker slips and breaks is an example of laboratory accident.

Wearing personal protective equipment will protect John.

In conclusion, following laboratory safety protocols will protect us when accidents occur in the laboratory.

Learn more about laboratory safety protocols at: brainly.com/question/17994387

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Note that the complete question is given as follows:

John is carefully pouring a chemical into a beaker when the beaker slips and breaks. How would laboratory safety protocols help John?

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1 year ago

I need part 1 and 2 please , just separate answers

Vladimir79 [104]

First, we have to remember the molarity formula:

$M=\text{ }\frac{moles\text{ of solute}}{L\text{ solution}}$

Part 1:

In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:

$M.W_{NaNO_3}=\text{ 23+14+16*3= 85 g/mol}$

Then, we calculate the moles present in the solution:

$3.976\text{ g NaNO}_3\text{ * }\frac{1\text{ mol}}{85\text{ g}}=\text{ 0.04678 mol NaNO}_3$

Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):

$M=\frac{0.04678\text{ mol}}{200\text{ mL*}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.2339 M}$

The molarity of this solution equals 0.2339 M.

Part 2:

In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:

$M=\text{ }\frac{0.04678\text{ mol}}{275\text{ mL *}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.1701 M}$

So, the molarity of this solution is 0.1701 M.

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