Answer:
4 hydrogen atoms are needed to C5H8 to saturate it
Explanation: because C5H8 is an alkyne that contains a triple bond or alkene with 2 double bonds.
2 hydrogen atoms are needed to saturate one double bond and 4 hydrogen atoms are needed to fully Saturate a triple bond or two double bonds.
The answer is [OH⁻] = 1 x 10⁻⁸.
To find OH⁻, divide the ionic product of water by [H₃O⁺] as :
<u>OH⁻ + H₃O⁺ = H₂O</u>
<u />
- [OH⁻] = 1 x 10⁻¹⁴ / 1 x 10⁻⁶
- [OH⁻] = 1 x 10⁻⁸
Q1)
firstly we need to determine the empirical formula of the compound. empirical formula is the simplest ratio of components in the compound.
percentages of the elements have been given, so lets assume we are calculating for a compound of 100g
C H O
mass 63.13 g 8.830 g 28.03 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.13/12 8.830/1 28.03/16
5.26 8.830 1.75
divide by the smallest number of moles
5.26/1.75 8.830/1.75 1.75/1.75
= 3.01 = 5.04 =1
rounded off to the nearest whole numbers
C - 3
H - 5
O - 1
therefore empirical formula = C₃H₅O
Q2)
we have to next determine the molecular formula of the compound
molecular formula gives the actual composition of elements in the compound.
since we know the empirical formula and molecular mass, we can find how many empirical units are in the molecular formula.
mass of empirical unit = Cx3 + Hx5 + Ox1
= 12 g/mol x 3 + 1g/mol x 5 + 16 g/mol x 1
= 36 + 5 + 16 = 57 g/mol
the molecular mass = 228 g/mol
then number of empirical units in the molecular formula = 228 / 57 = 4
therefore there are 4 empirical units
then the molecular formula = 4 x empirical formula =4 (C₃H₅O)
molecular formula = C₁₂H₂₀O₄
Answer: This is because, the stem symbolises the bronchioles of the lung through which oxygen is transfered to the lungs. The grabes represent the lungs, left and right respectively.
✩✩✩
Answer:
Option d.
Explanation:
Ketones contain a carbonyl groups as a functional group, which is a carbon bonded to oxygen with a double bond. In a ketone, the carbon is always bonded to two carbon atoms:
R-C(=O)-R'
The carbon in the carbonyl group has a hybridization sp2 (3 hybrid orbitals with an unhybridized p orbital), where two of the orbitals form sigma (σ) bonds with the other two carbons (R-C-R') and the other hybrid orbital form a sigma bond with the oxygen (C-O). The unhybridized p orbital on the carbon atom is used to form a pi (π) bond with the oxygen, thus forming the double bond (C=O).
The bond of a carbonyl group is polar, because of the difference of the electronegativity between the carbon atom and the oxygen atom.
Hence, from all of the above <u>we can discard the option a</u>, (the carbonyl groups exhibits sp2 hybridization), <u>the option b</u> (carbon-oxygen bond is a bond polar) and <u>the option c</u> (the group must always be in the middle of a carbon chain, the groups that are always in the end, are a aldehyde groups).
Therefore, the correct option is d, the functional group of this type of compound must always be in the middle of a carbon chain.
I hope it helps you!
