Answer: 0.002 M
Explanation:
The balanced chemical equation for ionization of in water is:
According to stoichiometry :
1 mole of ionizes to give =2 mole of
ions
0.001 mole of ionizes to give =
mole of
ions
Thus of a solution of 0.001 M aqueous sulfuric acid is 0.002 M
Answer:
Forms ionic bonds, insoluble, doesn't participate in single or double displacement reactions, non-reactive, high heat of combustion
Explanation:
Beryllium is a metal, since it belongs to group 2A, the alkaline earth metals. It has a total of 2 valence electrons.
Phosphorus, on the other hand, belongs to group 5A and has a total of 5 valence electrons.
We have a compound which has a metal in it, therefore, it's an ionic compound. Beryllium, our metal, loses its 2 electrons to gain an octet and phosphorus, our nonmetal, should gain 3 electrons to have an octet. The oxidation states are +2 and -3 respectively. This means we need 3 beryllium cations and 2 phosphide anions in our formula .
Beryllium phosphide would be expected to be insoluble, as only beryllium chloride, fluoride, nitrate, phosphate and sulfate are soluble substances, while the remaining ones are expected to be insoluble.
Due to its insolubility, beryllium phosphide would not participate in any ionic reactions, such as single displacement or double displacement.
Since it's insoluble, we expect this compound to be chemically stable and not reactive. This implies that if we wanted to burn it, the heat of combustion would be very high, as a lot of energy would be needed to be supplied in an endothermic reaction in order to burn it.
Answer:
None of these
Explanation:
Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.
Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :
Step -1 : Generation of stable carbocation.
Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.
Step-2: Attack of the ring to the carbocation
The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.
The product formed is shown in mechanism does not mention in any of the options.
So, None of these is the answer
