Answer:
magnesium oxide molecules
Explanation:
this is because the two is before the formula of MgO
Answer:
The mixture of cryolite and aluminum oxide has a lower melting point than pure aluminum oxide. This means a lower amount of energy is required to establish effective conditions for electrolysis and thus makes it more cost effective.
Explanation:
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.
Answer:
The halogen is Iodine.
Explanation:
Using the ideal gas equation, we find the number of moles of gas present, n.
PV = nRT where P = pressure of gas = 1.41 atm, V = volume of gas = 109 mL = 0.109 L, n = number of moles of gas, R = molar gas constant = 0.082 L-atm/mol-K and T = temperature of gas = 398 K
Since PV = nRT, making n subject of the formula, we have
n = PV/RT
substituting the values of the variables into the equation, we have
n = 1.41 atm × 0.109 L/(0.082 L-atm/mol-K × 398 K)
n = 0.15369 atm-L/32.636 L-atm/mol
n = 0.0047 mol
Since n = m/M where m = mass of gas = 1.19 g and M = relative molecular mass of gas
So, M = m/n
M = 1.19 g/0.0047 mol
M = 252.7 g
Since halogens are diatomic the relative atomic mass is M/2 = 252.7g/2 = 126.34 g
From tables, the only halogen with this atomic mass is Iodine.
So, the halogen is Iodine.
The answer is Mole, but send me a message if you need the equation for how to get it