Answer:
SbcI3
Explanation:
The symbol of antimony is 'Sb'.
The symbol of chlorine is 'Cl'
First write down the symbol of the first element.
Use the prefix to determine the atoms of first element. If there is no prefix on element then there is only 1 atom.
Now write down the symbol of the second element.
Use the prefix to determine the atoms of second element.
Use prefix as 'mono' for '1', 'di' for '2', 'tri' for '3' and so on.
Answer:
Oxygen's atomic weight is 16.00 amu. 1 mole of oxygen is 6.02 x 1023 atoms of oxygen 1 amu = 1.661 x 10-24g What is the molar mass (g/mole) of oxygen? Molar mass (in grams) is always equal to the atomic weight of the atom! Molar mass (in grams) is always equal to the atomic weight of the atom!
Answer:
Average atomic mass = 10.812 amu
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
<u>For first isotope, Boron-10:
</u>
% = 19.8 %
Mass = 10.013 amu
<u>For second isotope, Boron-11:
</u>
% = 80.2 %
Mass = 11.009 amu
Thus,
<u>Average atomic mass = 10.812 amu</u>
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.
