Answer:
0.00335 moles
Explanation:
From the question, Using
PV = nRT................... Equation 1
Where P = pressure, V = Volume, n = number of moles of argon gas, R = Molar gas constant, T = Temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: P = 1 atm (standard pressure), T = 273 K (standard temperature), V = 75 mL = 0.075 dm³
Constant: R = 0.082 atm·dm³/K·mol
Substitute into equation 2
n = (1×0.075)/(273×0.082)
n = 0.075/22.386
n = 0.00335 moles
Answer:
The Correct IUPAC name is H3C - CH (CH3) - CH (C2H5) - (CH2)2 - CH3 Class 11
Explanation:
yes searched np is maybe right i not 100% sure i 50% is it right >:) tell if u got it right >:D
Answer is: formula is Al₂(CO₃)₃.
Aluminium carbonate (Al₂(CO₃)₃) has neutral charge. Because aluminium cation has positive charge 3+ and carbonate anion has negative charge 2-, for right chemical formula, we need two aluminium cations and three carbonate anion:
charge of the molecule = 2 · (3+) + 3 · (-2).
charge of the molecule = 0.
Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Solution : Given,
Mass of Cu = 300 g
Molar mass of Cu = 63.546 g/mole
Molar mass of 
= 183.511 g/mole
- First we have to calculate the moles of Cu.
The moles of Cu = 4.7209 moles
From the given chemical formula, we conclude that the each mole of compound contain one mole of Cu.
So, The moles of Cu = Moles of = 4.4209 moles
- Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of = 4.4209 moles × 183.511 g/mole = 866.337 g
Mass of = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)
Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
