N=N₀*2^(-t/T)
N₀=200 g
T=10 d
t=30 d
N=200*2^(-30/10)=25 g
25 g will remain
E=hc/λ =6.626×10^-34×3 ×10^8 / 3×10^7 × 10^-9 = 6.626×10 ^-24J.
Answer:
75.15 mol.
Explanation:
- Firstly, we need to write the balanced equation of the reaction:
<em>Fe₂O₃ + 3CO → 2Fe + 3CO₂.</em>
It is clear that 1.0 mole of Fe₂O₃ reacts with 3.0 moles of CO to produce 2.0 moles of Fe and 3.0 moles of CO₂.
∴ Fe₂O₃ reacts with CO with (1: 3) molar ratio.
- we need to calculate the no. of moles of (4000 g) of Fe₂O₃:
<em>no. of moles of Fe₂O₃ = mass/molar mass</em> = (4000 g)/(159.69 g/mol) = <em>25.05 mol.</em>
<u>Using cross multiplication:</u>
1.0 mole of Fe₂O₃ needs → 3.0 moles of CO,
∴ 25.05 mole of Fe₂O₃ needs → ??? moles of CO.
<em>∴ The no. of moles of CO needed</em> = (3.0 mol)(25.05 mol)/(1.0 mol) =<em> 75.15 mol.</em>
I believe the answer is B. PO4-3
Answer:
the original answer is 38.9km (3sf)
