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Oksanka [162]
3 years ago
9

In a solution, the substance that changes phase is the _____, while the substance that does not change phase is the____.

Chemistry
1 answer:
Arisa [49]3 years ago
8 0
2) A) <span>solute, solvent

3) B) </span><span>Stirring mixes solute particles with solvent particles more thoroughly, providing more chances of interaction.

4) D) </span><span>concentrated and unsaturated

Hope this helps!</span>
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Tính axit giảm dần của các chất C6H5 OH (1) ,p – CH3 OC6H4 OH (2), p-NO2C6H4OH (3) pCH3COC6H4OH (4), p-CH3C6H4 OH (5) là:
nadya68 [22]

Answer:

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Explanation:

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3 0
2 years ago
30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr
Alenkasestr [34]

Answer: Thus concentration of Ca(OH)_2 in mol/dm^3  is 0.011 and in g/dm^3 is 0.814

Explanation:

To calculate the concentration of Ca(OH)_2, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3         1cm^3=0.001dm^3

Putting values in above equation, we get:

1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3

The concentration in g/dm^3 is 0.011mol/dm^3\times 74g/mol=0.814g/dm^3

Thus concentration of Ca(OH)_2 is 0.011mol/dm^3 and 0.814g/dm^3

4 0
3 years ago
How many moles of lithium nitrate<br> (LINO3) are in 256 mL of a 0.855 M<br> solution?
sweet [91]

Answer:

There are 0.219 mol of LINO3

7 0
3 years ago
Please help me no links <br> if good answer you get brainliest
Olin [163]
The crystals will not be oxidised since it will be underground
5 0
3 years ago
Read 2 more answers
Write the chemical equation for fot the reaction between mgcl2 and the soap you preapred
In-s [12.5K]
Soap is the sodium or potassium salt of long chain of fatty acid. Fatty acids when treated with NaOH or KOH forms Soap. This process is called as Saponification. Examples of Soap are as follow,

                                     1.  Sodium Stearate C₁₇H₃₅COONa
                                   
                                     2.  Potassium Oleate C₁₇H₃₃COOK

Reaction of Soap with MgCl₂;

When Soap is treated with MgCl₂ or CaCl₂ it forms insoluble precipitate called S.C.U.M. The reactions with MgCl₂ are as follow,

                2C₁₇H₃₅COONa + MgCl₂  -------->  2C₁₇H₃₅COOMg  + 2 NaCl

                2C₁₇H₃₃COOK   + MgCl₂  -------->  2C₁₇H₃₅COOMg  + 2 KCl

These reaction are often found in hard water. And this reaction decreases the effectiveness of soap.
5 0
3 years ago
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