No because carbon is a gas and sand is a solid
Answer: Weak acids are poor conductors of electricity.
Explanation:
Weak acids are defined as the acids which dissociate weakly or partially when dissolved in water (polar solvent).
For example, 
Due to less dissociation there will be less number of ions present in a solution of weak acid. As a result, a weak acid will be poor conductor of electricity.
As electricity is the flow of ions.
Thus, we can conclude that weak acids are poor conductors of electricity.
The statement that correctly summarizes the trend in electron affinity is the second option: it tends to be more negative across a period".<span> Electron affinity is the energy absorbed or released by a neutral atom when it absorbs an electron. The more negative the electron affinity the higher the affinity of the atom for an additional electron. So as the electronegativity trends to increase across a period, indicating that the atoms of the left side pull stronger the electrons, the fact that the electron affinity is more negative across a period tells that the atoms to the left accept better an additional electron.</span>
Answer:
The base is involved in the rate determining step of an E2 reaction mechanism
Explanation:
Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;
Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.
For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;
Rate = k[alkyl halide] [base]
The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.
Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.
54.0 ml of a 1.2 m solution was diluted to a total volume of 228 ml
USing dilution equation
M1V1 = M2V2
Where M1V1 is before dilution while M2V2 is after dilution.
Therefore;
M1= 1.20 M, V2= 54 ml,
M2= V2 = 228 ml
1.20 M × 54 ml = M2 × 228 ml
M2 = (1.2 ×54)/ 228 ml
= 0.2842 M
0.2842 × 114 ml = M2 × (114 +111)
M2 = (0.2842 ×114)/ 225
= 0.143995
= 0.144 M
