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Archy [21]
3 years ago
8

Change the coefficients until the reaction is balanced. (also if anyone could help me find the rest of the gizmos I would be ver

y appreciative lol)

Chemistry
1 answer:
Artemon [7]3 years ago
6 0

Answer:

1Na₂S + 1Cd(C₂H₃O₂)₂ →  1CdS + 2NaC₂H₃O

Explanation:

1Na₂S + 1Cd(C₂H₃O₂)₂ → 1CdS + 1NaC₂H₃O₂

1Na₂S + 1Cd(C₂H₃O₂)₂ → 1CdS + 2NaC₂H₃O₂

(on the reactant side there are two molecules of C₂H₃O₂ and 2 atoms of Na, so in order to balance that we put the coefficient 2 on NAC₂H₃O₂)

So the final product is

1Na₂S + 1Cd(C₂H₃O₂)₂ →  1CdS + 2NaC₂H₃O₂

You might be interested in
NiS2(s) + O2(g) --> NiO(s) + SO2(g) When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained. The theoretical
makkiz [27]

Answer:

1. The theoretical yield of NiO is 5.09g.

2. O2 is the limiting reactant.

3. The percentage yield of NiO is 95.5%

Explanation:

Step 1:

The balanced equation for the reaction is given below:

2NiS2(s) + 5O2(g) —> 2NiO(s) + 4SO2(g)

Step 2:

Determination of the masses of NiS2 and O2 that reacted and the mass of NiO produced from the balanced equation. This is illustrated below below:

Molar mass of NiS2 = 59 + (32x2) = 123g/mol

Mass of NiS2 from the balanced equation = 2 x 123 = 246g

Molar mass of o3= 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar mass of NiO = 59 + 16 = 75g/mol

Mass of NiO from the balanced equation = 2 x 75 = 150g

Summary:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2 to produce 150g of NiO

Step 3:

Determination of the limiting reactant. This can be obtain as follow:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2.

Therefore, 11.2g of NiS2 will react with = (11.2 x 160)/246 = 7.28g of O2.

From the above calculation, we can see that it will take a higher mass of O2 i.e 7.28g than what was given i.e 5.43g to react completely with 11.2g of NiS2.

Therefore, O2 is the limiting reactant and NiS2 is the excess reactant.

1. Determination of the theoretical yield of NiO.

In this case, the limiting reactant will be used as all of it is consumed in the reaction. The limiting reactant is O2.

From the balanced equation above, 160g of O2 reacted to produce 150g of NiO.

Therefore, 5.43g of O2 will react to produce = (5.43 x 150)/160 = 5.09g of NiO.

Therefore, the theoretical yield of NiO is 5.09g.

2. The limiting reactant is O2. Please review step 3 above for explanation.

3. Determination of the percentage yield of NiO. This is illustrated below:

Actual yield of NiO = 4.86g

Theoretical yield of NiO = 5.09g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 4.86/5.09 x 100

Percentage yield of NiO = 95.5%

3 0
3 years ago
What is e° cell for the reaction of nickel(ii) ions with cadmium metal at 25°c? ( r = 8. 314 j/k • mol, f = 96,500 c • mol –1) n
Alexus [3.1K]

At anode,

Cd----->  + 2

At cathode,

 + 2e -----> Ni

Overall reaction,

+ Cd(s) ------> + Ni                          K=1.17

(aq)                          (aq)      (s)

By nernst equation,

E=  -  Log K

Where,

          F=96500c/mol

         M=2

          T=298K

         E=Zero at equilibrium

R=8.314 J

0=  - Log(1.17)

= 0.150v

what is nernst equation?

The Nernst equation is a chemical thermodynamical relationship used in electrochemistry that enables the determination of a reaction's reduction potential (half-cell or full-cell reaction) from the standard electrode potential, absolute temperature, the number of electrons involved in the oxydo-reduction reaction, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation, respectively. It was given the equation's original name in honor of the German physical scientist Walther Nernst.

The link between cell potential, standard potential, and the behaviors of electrically active (electroactive) species is described by the Nernst equation. It links the standard cell potential to the effective concentrations (activities) of the reaction's constituent parts.

Learn more about nernst equation visit here:-

brainly.com/question/75084?referrer=searchResults

#SPJ4

5 0
2 years ago
What is change in atomic number when an atom emits positron?
USPshnik [31]
When a nucleus emits a positron, its atomic number is lowered by one but its mass number stays the same.
8 0
4 years ago
Read 2 more answers
Calculate the mass of chlorine in a 20g tablet of calcium hypochlorite​
Kobotan [32]

Answer:

              9.92 g of Cl

Explanation:

Well first find out the molar mass of calcium hypochlorite.

The molecular formula is,

                                          CaO₂Cl₂

Adding the atomic masses we get the molar mass = 142.98 g/mol

This means,

                                 142.98 f of CaO₂Cl₂ = 70.91 g of Cl

Therefore,

                                       20 g of CaO₂Cl₂ = X

Solving for X,

               X = (70.91 g of Cl × 20 g of CaO₂Cl₂) ÷ 142.98 f of CaO₂Cl₂

               X = 9.92 g of Cl

3 0
3 years ago
In your own words, explain why each of the filler blocks (s,p,d,f) on the periodic chart have their characteristic width(number)
Charra [1.4K]

Answer:

Each of the four blocks of the periodic table is the collection of elements that have common valence orbitals, or that their valence are found in one of the four orbitals with azimuthal quantum number 0, 1, 2, or 3 which corresponds to the shape (s), principal (p), diffuse (d), or fundamental (f) orbitals respectively

The maximum number of electrons in the s-orbital = 2 electrons

The maximum number of electrons in the p-orbital = 6 electrons

The maximum number of electrons in the d-block = 10 electrons

The maximum number of electrons in the f-orbital = 14 electrons

Therefore, given that the elements of the within each of the blocks have their valence electron in the corresponding orbital in which they are located on the periodic table, given the periodic nature of the periodic table, the width of each block of elements corresponds to the maximum number of elements that can be found in their valence orbital.

Therefore, we have;

The width of the s-block elements = 2 block units

The width of the p-block elements = 6 block units

The width of the d-block elements = 10 block units

The width of the f-block elements = 14 block units

Explanation:

6 0
3 years ago
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