Answer is: identity of the metal is gold (Au).
ω(Cl) = 35.06% ÷ 100%.
ω(Cl) = 0.3506; mass percentage of chlorine.
If we take 100 grams of the compound:
m(Cl) = ω(Cl) · m(compound).
ω(Cl) = 0.3506 · 100 g.
ω(Cl) = 35.06 g.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 35.06 g ÷ 35.45 g/mol.
n(Cl) = 0.99 mol; amount of substance.
In molecule MCl₃: n(M) : n(Cl) = 1 : 3.
n(M) = 0.33 mol; amount of unknown metal.
M(M) = m(M) ÷ n(M).
M(M) = (100 g - 35.06 g) ÷ 0.33 mol.
M(M) = 196.8 g/mol; molar mass of the gold.
Answer: the mass of the object
Explanation: To find density, the equation d= m/v is needed where m= mass and v= volume
The name is Copper(II) Oxide and/or Cupric Oxide.
Cu is Copper and O is Oxygen.
Usually when the Oxygen ion is placed after another, it's name is Oxide.
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>:
