Answer:
ΔT = Tfinal − Tinitial = 150°C − 35.0°C = 125°C
given the specific heat of iron as 0.108 cal/g·°C
heat=(100.0 g)(0.108 cal /g· °C )(125°C) =
100x 0.108x125= 1350 cal
First, we write the balanced equation for this reaction:
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:
Moles = volume (in L) * molarity
We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:
M₁V₁ = 2M₂V₂
V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L
The volume required is 173.6 mL
Answer:
The answer is 0.023 moles of phosphorus
Explanation:
The 15-15-15 fertilizer is a fertilizer of great versatility, made with nitrogen, phosphorus and potassium, which makes it one of the fertilizers most used for fertilizer in the sowing plant, thus covering the crop requirements from planting. .
This fertilizer consists of 14.25% phosphorus pentoxide (P2O5). Therefore, we have to remove 14.25% at 10 grams of 15-15-15 fertilizer to calculate the moles of phosphorus. As follows:
Grams of P2O5 = 10 g x 0.1425 = 1.425 g
We calculate the molecular weight of phosphorus. We use the periodic table:
Phosphorus molecular weight = 2 x 30.97 = 61.94 g/mol
Now we calculate the moles of phosphorus in the fertilizer:
Phosphorus moles = 1,425 g/61.94 g/mol = 0.023 moles
