<span>To find the mass of 3.00 moles of magnesium chloride (MgCl2), first record the atomic mass of magnesium (Mg) and chloride (Cl), which are both listed on the periodic table as follows:
Mg=24 g/mole
Cl=38 g/mole
Now, double the Cl mass since there are 2 Cl moles in MgCl2 and then add it to the Mg mass like so:
(38 g/mole*2 moles)+24 g/mole=100 g/mole
Finally, to calculate the mass of 3.00 moles of MgCl2, convert the combined atomic mass to grams as follows:
3.00 moles * 100 g/mole = 300 g</span>
Answer:
1.1 mol
Explanation:
n=m/M, where n is moles, m is mass, and M is molar mass.
M of CO2 = 12.01+16.00+16.00 = 44.01g/mol
n=50g/44.01g/mol
n = 1.13610543 mol
n ≈ 1.1 mol
Hope that helps
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
Answer:
The smallest unit of a compound is a molecule, which is made up of atoms held together by bonds.
