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Dafna11 [192]
2 years ago
14

Why was the measurement unit for metal content in Experiments 1 and 2 different? In Experiment 1:

Chemistry
1 answer:
olya-2409 [2.1K]2 years ago
8 0

To identify why a metal measurement was different in the experiments look for the variable that was different in the experiment and analyze how this change affected the results.

<h3>What is an experiment?</h3>

An experiment is a procedure that aims at probing or discovering something. For example, you can test if a plant grows faster/slower by using an experiment.

<h3>What causes different results in similar experiments?</h3>

The most common cause for this situation is that one of the factors or variables is slightly different. For example, if I add 50mL of water to a plant rather than 20mL of water every day this might cause different results.

Based on this, if the metal content was different you should analyze if any of the factors changed in this experiment and find out how this change affected the general results.

Note: This question is incomplete because there is limited information about the experiment; due to this, I answered it based on general knowledge.

Learn more about experiments in: brainly.com/question/13270830

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Explain why the boiling point of H2S with relative molecular mass 34 is lower than that of H2O with relative molecular mass 18​
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8 0
3 years ago
If 4.12 l of a 0.850 m-h3po4 solution is be diluted to a volume of 10.00 l, what is the concentration of the resulting solution?
statuscvo [17]

The process in which the concentration of the solution is lessened  by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.

Formula is given by:

C_{1}V_{1}=C_{2}V_{2}   (1)

where,

C_{1} is the initial concentration

V_{1} is the initial volume

C_{2} is the final concentration

V_{2} is the final volume

Now,

C_{1} = 0.850 M

V_{1} = 4.12 L

C_{2}  =?

V_{2} = 10.00 L

Substitute the give values in formula (1),

0.850 M\times 4.12L=C_{2}\times 10.00 L

C_{2} =\frac{0.850 M\times 4.12L}{10.00 L}

= 0.3502 M

Thus, the final concentration of theH_{3}PO_{4} solution = 0.3502 M












5 0
2 years ago
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