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valentinak56 [21]
2 years ago
9

When aqueous solutions of NH4OH(aq) and CuCl2(aq) are mixed, the products are NH4Cl(aq) and Cu(OH)2(s). What is the net ionic eq

uation for this reaction
Chemistry
1 answer:
Ilia_Sergeevich [38]2 years ago
5 0

Answer:

2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)

Explanation:

The net ionic equation usually shows the main ionic reaction that goes in the system. The other ions that do not participate in this net ionic equation are called spectator ions. Spectator ions do not participate in the main reaction occurring in the system.

The net ionic equation quite often result in the formation of a solid precipitate in the system such as Cu(OH)2.

The net ionic equation for this reaction is;

2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)

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A chemist adds 180.0 mL of a 1.42M sodium carbonate (Na CO,) solution to a reaction flask. Calculate the millimoles of sodium ca
svet-max [94.6K]

Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256

Explanation:

Molarity is defined as the number of moles dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{milli moles of solute}}{\text{Volume of solution in ml}}     .....(1)

Molarity of BaCl_2 solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in equation 1, we get:

1.42M=\frac{\text{milli moles of }BaCl_2}{180.0ml}\\\\\text{milli moles of }BaCl_2}={1.42M\times 180.0ml}=256milli mol

Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.

6 0
2 years ago
An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th
melisa1 [442]
570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students
7 0
2 years ago
Read 2 more answers
The Rio Grande river flows through the South Texas Plains ecoregion of Texas. The river is a major source of water for cities an
mylen [45]

Answer:

What are the answer choices

Explanation:

7 0
1 year ago
Read 2 more answers
If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
2 years ago
How many atoms of zirconium are in 0.3521 mol of zirconium?
lora16 [44]

Answer:

2.12×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.

Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:

1 mole of zirconium also 6.02×10²³ atoms.

Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.

Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.

3 0
2 years ago
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