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Leni [432]
3 years ago
5

Two unknown household items are being tested with litmus paper. Substance A turns red litmus paper blue and Substance B turns bl

ue litmus paper red. Which of the following best explains the results of the experiment?
Both Substances A and B are bases.

Substance A is an acid and Substance B is a base.

Both Substances A and B are acids.

Substance A is a base and Substance B is an acid.
Chemistry
1 answer:
irina1246 [14]3 years ago
7 0

Answer:

Substance A is a base and Substance B is an acid.

According to the experimental results

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Please help me with chemistry
vodka [1.7K]

Table Giving Answer

Element Atomic mass % Amount

Mg_24     24                79    18.96

Mg_25 25                10    2.5

Mg_26 26                11   2.86

   

Total                     24.32

Discussion

The method of calculation for this table, which was done in Excel (a spread sheet) is shown below. Assume that there is 100 grams of material of "pure" magnesium. What is it's mass?

<em><u>Sample  Calculation</u></em>

The the sample atomic mass = 24

Mass = % * sample atomic mass

Mass = 79% * 24

Mass = (79/100) * 24

Mass = 18.96

<em><u>Note</u></em>

The other two elements are found exactly the same as the sample calculation.

Then all you do is add the 3 masses together.

Answer

The mass of Mg to 1 decimal place is 24.3 <<<< Answer.


7 0
3 years ago
If you combine 27.1 g of a solute that has a molar mass of 27.1 g/mol with 100.0 g of a solvent, what is the molality of the res
baherus [9]

<u>Answer:</u> The molality of the solution is 0.1 m.

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute = 27.1 g

M_{solute} = Molar mass of solute = 27.1 g/mol

W_{solvent} = Mass of solvent = 100 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{27.1\times 1000}{27.1\times 100}\\\\\text{Molality of solution}=0.1m

Hence, the molality of the solution is 0.1 m.

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