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:
1.58x10⁻⁵
2.51x10⁻⁸
0.0126
63.10
Explanation:
Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:
pH = pKa + log[In-]/[HIn]
pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,
i) pH = 4.9
4.9 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = - 4.8
[In-]/[HIn] = 
[In-]/[HIn] = 1.58x10⁻⁵
ii) pH = 2.1
2.1 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -7.6
[In-]/[HIn] = 
[In-]/[HIn] = 2.51x10⁻⁸
iii) pH = 7.8
7.8 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -1.9
[In-]/[HIn] = 
[In-]/[HIn] = 0.0126
iv) pH = 11.5
11.5 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = 1.8
[In-]/[HIn] = 
[In-]/[HIn] = 63.10
Use a strip of paper covered in PH indicating dye.
Answer:
it says select all that apply hhh easy just select and apply the answer hhh
Explanation:
Answer:
The mass percentage of carbon can be found easily using the molar mass of C6H12O6, 180.1559 g/mol. We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass. C6H12O6 = CO2 + C3H6O3 + CH3OCH3 Take fructose for example. Compound.
Explanation: I looked it up
