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:
3853 g
Step-by-step explanation:
M_r: 107.87
16Ag + S₈ ⟶ 8Ag₂S; ΔH°f = -31.8 kJ·mol⁻¹
1. Calculate the moles of Ag₂S
Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S
2. Calculate the moles of Ag
Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag
3. Calculate the mass of Ag
Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag
You must react 3853 g of Ag to produce 567.9 kJ of heat
Answer: In simplest case mass of reactants is same as mass of products.
Without thinking this question deeper, mass of ZnCl2 would be 49, but..
Explanation: Reaction should be Zn + 2 HCl ⇒ ZnCl2 + H2
Amount of zinc is 5 g / 65,38 g/mol = 0,076476 mol and amount
of Hydrogen Chloride is 50 g / 36.458 g/mol = 1,371 mol.
Althought HCl is needed 0.152 moles, zinc is an limiting reactant.
So it is possible to produce only 0.076476 mol Hydrogen and its mass
is 0.154 g. Mass of ZnCl2 would be 0.076476 mol · (65.38 + 2·35.45) =
10.42 g
