Answer:
2) Copper (II) Chloride
Explanation:
A precipitate will form if the resulting compound is insoluble in water. For example, a silver nitrate solution (AgNO3) is mixed with a solution of magnesium bromide (MgBr2).
Answer:
Y is a 3-chloro-3-methylpentane.
The structure is shown in the figure attached.
Explanation:
The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).
The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.
the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g
The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn
3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2
the formula is n= mass/M so, now substituting values
m (Al2O3)= m(MnO2) X 2 X M (Al2O3) / M(MnO2 X3
so, by substituting values, 2 X101.96 X1.74g / 3 X 86.94 =1.36g
so mass of aluminum oxide obtained = 1.36g
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Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:
Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:
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