Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O
With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
Answer:
4.67M
Explanation:
The concentration of methanol (CH3OH) can be calculated using the following:
Molarity (M) = number of moles(n)/volume(v)
However, mole is not given. It can be obtained by using:
Mole = mass / molar mass
Where; mass = 34.4g
Molar mass (MM) of CH3OH is:
= 12 + 1(3) + 16 + 1
= 12 + 3 + 17
= 32g/mol
mole = 34.4/32
mole = 1.075mol
The volume needs to be converted to L by dividing by 1000
230mL = 230/1000
= 0.230L
Molarity = mol/volume
Molarity = 1.075/0.230
Molarity = 4.6739
Molarity = 4.67M
The concentration of CH3OH in solution is 4.67M
Answer:
C. Y & Z
Explanation:
V, W are imaginary metals here because their valence electrons are typically less than 4. X, Y, Z are non-metals and have higher valence electrons. Here, if V or W bind with X, Y, or Z we make ionic bond (because metal + non metal = ionic). But, if X binds with Y or Z or any combinations of any two of the three non-metals results in covalent bond (non metal + non metal = covalent).
Thus, Y and Z make covalent.
