Answer:
227.78g of the precipitate are produced
Explanation:
Based on the reaction, 3 moles of CuCl2 produce 1 mole of Cu3(PO4)2 (The precipitate).
To solve this question we need to find the moles of CuCl2 added. With these moles and the reactio we can find the moles of Cu3(PO4)2 and its mass as follows:
<em>Moles CuCl2:</em>
285mL = 0.285L * (6.3mol / L) = 1.7955 moles CuCl2
<em>Moles Cu3(PO4)2:</em>
1.7955 moles CuCl2 * (1mol Cu3(PO4)2 / 3mol CuCl2) = 0.5985 moles Cu3(PO4)2
<em>Mass Cu3(PO4)2 -380.58g/mol-</em>
0.5985 moles Cu3(PO4)2 * (380.58g/mol) =
227.78g of the precipitate are produced
Answer: P2O5 is the empirical formula.
Explanation: When given percentages you can assume that many grams of each atom are in the compound. Then you divide grams by the molar mass of each element, giving you moles. Once you have moles, divide by the smaller molar amount, which should give you 1 mol of Phosphorus and 2.5 mol of Oxygen. Then multiply by 2 in order for both moles to be a whole number. This gets you 2 and 5.
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
