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Sophie [7]
4 years ago
9

What is a chemical property of soda ash

Chemistry
1 answer:
vovikov84 [41]4 years ago
8 0

A chemical sodium carbonate (Na2CO3), produced by reacting sodium chloride and limestone, is commonly known as soda ash.

Soda ash has a number of chemical properties and one of the important properties is that it a basic i.e. an alkaline compound. It can be used as a pH regulator for a number of processes such as raising the pH of a swimming pool.

It is also a good electrolyte and can make a good conductive solution when add to water.


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4 0
3 years ago
Read 2 more answers
How many grams of lithium hypochlorite (LiClO) are there in 0.594 moles?
Flauer [41]

Answer : The mass of lithium hypochlorite are, 34.7 grams.

Explanation : Given,

Moles of LiClO = 0.594 g

Molar mass of LiClO = 58.4 g/mol

Expression used :

\text{ Mass of }LiClO=\text{ Moles of }LiClO\times \text{ Molar mass of }LiClO

Now put all the given values in this expression, we get:

\text{ Mass of }LiClO=(0.594moles)\times (58.4g/mole)

\text{ Mass of }LiClO=34.7g

Therefore, the mass of lithium hypochlorite are, 34.7 grams.

3 0
3 years ago
Explain why lime is added to lakes. Refer to the problem that exists and what lime does to the lakes.
aalyn [17]

Explanation:

Lime is added to lower the acidity of lakes when pH levels go too low.

Acidic environment is harmful to aquatic ecosystems. Acidity causes some minerals, such as phosphorus and calcium, in the water to become unavailable to aquatic organisms like plants and fishes ( This means the water hardness is reduced). The low nutrients cause stress and death in the aquatic environment lowering biodiversity.  Applying lime raises the pH to neutral levels and improves the health of the aquatic life in the lake. This also improves economic activities in the lake like fishing.

Learn More:

For more on acidicity of lakes check out;

brainly.com/question/3612253

brainly.com/question/3612253

#LearnWithBrainly

4 0
3 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
3 years ago
If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density
adelina 88 [10]

Explanation:

The given data is as follows.

      Mass = 27.9 g/mol

As we know that according to Avogadro's number there are 6.023 \times 10^{26} atom present in 1 mole. Therefore, weight of 1 atom will be as follows.

            1 atoms weight = \frac{38}{6.023 \times 10^{26}}    

In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.

Therefore, total weight of atoms in a unit cell will be as follows.

            = \frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}

            = 37.06 \times 10^{-26}

Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.

                   = a^{3}

                   = (0.503 \times 10^{-9})^{3}

                   = 0.127 \times 10^{-27} m^{3}

Formula to calculate density of diamond cell is as follows.

               Density = \frac{mass}{volume}

                             = \frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}

                            = 2918.1 g/m^{3}

or,                         = 0.0029 g/cc       (as 1 m^{3} = 10^{6} cm^{3})

Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.

4 0
3 years ago
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