Answer : The three-dimensional structure of ethanol is shown below.
Explanation :
The ethanol molecule is made up of 2 carbon atoms, 6 hydrogen atoms and 1 oxygen atom.
The 3D representation is shown by solid, dash and wedge bonds.
Dash lines shows that the molecules are behind the plane.
Wedge shows that the molecules are above or front of the plane.
Solid line shows that the molecules are present on the plane.
The three-dimensional structure of ethanol is shown below.
Answer:
54.1 % Ca, 43.2 % O, 2.7% H
Explanation:
Molecular formula for calcium hydroxide is Ca(OH)₂
As we don't have a mass of Ca(OH)₂ to find out the percentage composition, we consider that the question refers to 1 mol of compound.
1 mol of hydroxide weighs 74.08 g
1 mol of hydroxide has 1 mol of Ca, therefore 40.08 g are Ca
2 moles of O therefore 32g are O
2 moles of H therefore 2 g are H
Percentage composition is known as (Mass of element/Total mass) . 100
- (40.08 / 74.08) . 100 = 54.1 %
- (32 / 74.08) . 100 = 43.2 %
- (2 / 74.08) . 100 = 2.7%
Answer:
after 45 days 9 g left
Explanation:
Given data:
Half life Na-24 = 15 days
Mass of sample = 72 g
Mass remain after 45 days = ?
Solution:
Number of half lives passed:
Number of half lives = time elapsed / half life
Number of half lives = 45 days / 15 days
Number of half lives = 3
At time zero total amount = 72 g
After first half life = 72 g/ 2= 36 g
At 2nd half life = 36 g/2 = 18 g
At 3rd half life = 18 g/2 = 9 g
Thus after 45 days 9 g left.
Answer: 2 H2 + O2 → 2 H2O, you get the same number of moles of water as H2, as long as you have 1 mole of O2. So, with 3 moles of H2, as long as you have 1.5 moles of O2, you will get 3 moles of H2O. In my opinion.
Find the mass of C in the 2.657 g CO2:
(2.657 g CO2) / (44.01 g/mol) = 0.06037 mol CO2
Since each mole of CO2 also has 1 mole of C, this is equivalent to 0.06037 mol C.
Find the mass of H in the 1.089 g H2O:
(1.089 g H2O) / (18.02 g/mol) = 0.06043 mol H2O
Since 1 mol H2O has 2 mol H, this is equivalent to (0.06043)*2 = 0.1209 mol H.
Taking the ratio of H to C: 0.1209 / 0.06037 = 2.002 ~ 2
Therefore, the empirical formula of isobutylene is CH2.
