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Paladinen [302]
3 years ago
7

A student, wearing chemical safety goggles and a lab apron, is to perform a laboratory test to determine the pH value of two dif

ferent solutions. The student is given one bottle containing a solution with a pH of 2.0 and another bottle containing a solution with a pH of 5.0. The student is also given six dropping bottles, each containing a different indicator listed in Reference Table M.
State one safety precaution, not mentioned in the passage, that the student should take while performing tests on the samples from the bottles.
Chemistry
1 answer:
Setler79 [48]3 years ago
6 0
They should probably wear gloves. A pH of 2.0 is a very strong acid, and it can easily irritate their hands. They should also remember to NEVER touch their face, mouth, etc. and to keep all chemicals on the laboratory table.
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Igoryamba

Answer:

The arm that was not sprayed with anything

Explanation:

The control group would be <u>the arm that was not sprayed with anything</u>.

<em>The control group during an experiment is a group that forms the baseline for comparison in other to determine the effects of a treatment. The control group does not include the variable that is being tested and as such, it provides the benchmark to measure the effects of the tested variable on the other group - the experimental group. In this case, the experimental group would be the arm that was sprayed with the repellent.</em>

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What are the atoms of Fe(SCN)3
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Explanation:

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Westkost [7]

Answer:

Law of conservation of mass

Ernest Rutherford

Explanation:

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This is the law of conservation of mass. It is very essential in understanding most chemical reaction. Also, in quantitative analysis, this law is pivotal.

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4 0
3 years ago
Determine the volume of water to be added to the nitric acid solution at a concentration of 8.61 mol / L to prepare 500 mL of th
Alex_Xolod [135]

Answer:

398 mL

Explanation:

Using the equation for molarity,

C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L

V₂ = V₁ + V' where V' = volume of water added.

So, From C₁V₁ = C₂V₂

V₁ =  C₂V₂/C₁

= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L

= 0.875 mol/8.61 mol/L

= 0.102 L

So, V₂ = V₁ + V'

0.5 L = 0.102 L + V'

V' = 0.5 L - 0.102 L

= 0.398 L

= 398 mL

So, we need to add 398 mL of water to the nitric solution.

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