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?
Answer:
The density of the ideal gas is directly proportional to its molar mass.
Explanation:
Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.
......equation (1)
According to the ideal gas law for ideal gas:
......equation (2)
Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas
As we know,
The number of moles: 
where m is the given mass of gas and M is the molar mass of the gas
So equation (2) can be written as:

⇒ 
⇒
......equation (3)
Now from equation (1) and (3), we get
⇒ Density of an ideal gas:
⇒ <em>Density of an ideal gas: ρ ∝ molar mass of gas: M</em>
<u>Therefore, the density of the ideal gas is directly proportional to its molar mass. </u>
The change in temperature (ΔT) : 56.14 ° C
<h3>Further explanation</h3>
Given
Cereal energy = 235,000 J
mass of water = 1000 g
Required
the change in temperature (ΔT)
Solution
Heat can be formulated :
Q = m . c . ΔT
c = specific heat for water = 4.186 J / gram ° C
235000 = 1000 . 4.186 . ΔT
