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stira [4]
3 years ago
13

Which one of these salts will form a basic solution upon dissolving in water?

Chemistry
1 answer:
daser333 [38]3 years ago
6 0
Https://www.numerade.com/ask/question/which-one-of-these-salts-will-form-a-basic-solution-upon-dissolving-in-water-licl-nh4no3-kbr-alcl3-nano2-04685/
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The atomic number of lithium is 3. How many neutrons does an atom of lithium have if it is represented by the symbol shown below
Mademuasel [1]
If it is not an isotope then it has around 21 neutrons
5 0
3 years ago
In the late 1400s,ties among people of the same formed the basics of most aspects of life in rural west Africa
lara31 [8.8K]
Can u write it more specifically? so that I can answer
6 0
4 years ago
35. In the collision theory, a collision that leads to the formation of products is called an
FinnZ [79.3K]

Answer:

It's Effective Collision.

Explanation:

Hope my answer has helped you!

7 0
3 years ago
13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What
rjkz [21]

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
6 0
3 years ago
A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of
blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of H^+ ion = Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol

Moles of OH^- ion = Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol

HCl+NaOH\rightarrow NaCl+H_2O

For neutralization:

1 mole of H^+ ion will react with 1 mole of OH^- ion

0.01 mol of H^+  ion will react with =\frac{1}{1}\times 0.01mole of OH^- ion

Thus (0.012-0.01)= 0.002 moles of OH^- are left in 20 ml or 0.02 L of solution.

[OH^-]=\frac{0.002}{0.02L}=0.1M

pOH=-log[OH^-]

pOH=-log[0.1]=1

pH+pOH=14

pH=14-1=13

Thus the pH of resulting solution will be 13

7 0
3 years ago
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