So yield = actual/predicted*100
You need to find the theoretical yield. CONVERT INTO MOLES. this is the 1st step you do in most calculation questions.
I got 15.92%
<h3>
Answer:</h3>
Ni + Pb(NO₃)₂ → Ni(NO₃)₃ + Pb
<h3>
Explanation:</h3>
We are required to write a balanced equation from the word equation;
- Nickel reacts with lead nitrate (II) to produce nickel(III) nitrate and lead
- The equation will be written by writing the symbols of the reactants and products.
That is;
Ni + Pb(NO₃)₂ → Ni(NO₃)₃ + Pb
We then balance the equation;
- To balance the equation, we put appropriate coefficients on reactants and products, so that the number of atoms of each element is equal on both sides of the equation.
- Thus, the balanced equation will be;
2Ni + 3Pb(NO₃)₂ → 2Ni(NO₃)₃ + 3Pb
Answer:
Percentage yield = 30%
Explanation:
Given data:
Number of moles of NO = 7.0 mol
Number of moles of O₂ = 5 mol
Number of moles of NO₂ = 3 mol
Percentage yield = ?
Solution:
Chemical equation:
2NO + O₂ → 2NO₂
Now we will compare the moles of NO₂ with NO and O₂ .
NO : NO₂
2 : 2
7.0 : 7.0
O₂ : NO₂
1 : 2
5.0 : 2 ×5.0 = 10 mol
The number of moles of NO₂ produced by NO are less it will be limiting reactant.
Mass of NO₂ = moles × molar mass
Mass of NO₂ = 10 mol × 46g/mol
Mass of NO₂ = 460 g
Actual yield of NO₂:
Mass of NO₂ = moles × molar mass
Mass of NO₂ = 3 mol × 46g/mol
Mass of NO₂ = 138 g
Percentage yield:
Percentage yield = Actual yield/theoretical yield × 100
Percentage yield = 138 g/ 460 g × 100
Percentage yield = 30%
The answer would be C there would be 6 nonbonding electrons and 2 bonding electrons.
Answer:
235/92U+10n→144/54Xe+90/38Sr+2/10n
Explanation:
- The nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90 is represented by;
235/92U+10n→144/54Xe+90/38Sr+2/10n
- In nuclear fission reactions a heavy nuclide is split into two light nuclides and is coupled by the release of energy.
