S + O2 → SO2
<span>z / (32.0655 g S/mol) x (1 mol SO2 / 1 mol S) x (64.0638 g SO2/mol) = (1.9979 z) g SO2 </span>
<span>C + O2 → CO2 </span>
<span>(9.0-z) / (12.01078 g C/mol) x (1 mol CO2 / 1 mol C) x (44.00964 g CO2/mol) = (32.9776 - 3.66418 z) g CO2 </span>
<span>Add the two masses of SO2 and CO2 and set them equal to the amount given in the problem: </span>
<span>(1.9979 z) + (32.9776 - 3.66418 z) = 27.9 </span>
<span>Solve for z algebraically: </span>
<span>z = 3.0 g S</span>
Answer:
329.7%
Explanation:
Percent Yield = Actual Yield/ Theoretical Yield x 100%
Percent Yield = 105.5g/32 x 100% = 329.69 ≈ 329.7 %
Answer:
The orbital shapes are actually representation of (Ψ)2 all over the orbit simplified ... ψnlml(r,θ,ϕ)=Rnl(r)Ymll(θ,ϕ) , ... and thus it is directly linked to the angular and radial nodes. ... for different quantum values(which can be assigned to different orbitals are ) .... The two types of nodes are angular and radial.
Explanation:
hope it helps
The reaction equation is:
<span>2CuO(s) + C(s) </span>→ <span>2Cu(s) + CO</span>₂<span>(g)
First, we determine the number of grams present in one ton of copper oxide. This is:
1 ton = 9.09 x 10</span>⁵ g
We convert this into moles by dividing by the molecular mass of copper oxide, which is:
9.09 x 10⁵ / 79.5 = 11,434 moles
Each mole of carbon reduces two moles of copper oxide, so the moles of carbon required are:
11,434 / 2 = 5,717 moles of Carbon required
The mass of carbon is then:
5,717 x 12 = 68,604 grams
The mass of coke is:
68,604 / 0.95 = 72,214 g
The mass of coke required is 7.22 x 10⁴ grams
Answer:
All answers attached in the pictures above.
