Answer:
B. Cu (s) +Ni(NO3)2 (aq) - 2CuNO3 +Ni (s)
Explanation:
the above reaction is a substitution reaction
In order to find the molarity of the solution, we first require the moles of acetic acid added. For this,we need the mass which is:
Mass = volume * density
Mass = 50 * 1.05
Mass = 52.5 grams
Moles = mass / molecular weight
Moles = 52.5 / 60.05
Moles = 0.874 mol
Next, we know that the molarity of a solution is:
Molarity = moles / liter
Molarity = 0.874 / 0.5
Molarity = 1.75 M
The question is incomplete, the complete question is;
AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the formula AlBr3 is —
aluminum bromide
monoaluminum tribromide
aluminide bromine
aluminum tribromide
Answer:
aluminum bromide
Explanation:
Having known that AlBr3 is an ionic compound and aluminium is the central atom here, we now have to ask ourselves if Aluminium exists in other stable oxidation states.
We must take cognizance of the fact that the oxidation number of the central atom in a compound becomes part of the name of that compound when other stable oxidation states for atoms of the same elements exists.
Since the +3 state is the only stable oxidation state for aluminium, the name of the compound is simply aluminium bromide.
Answer:
Percent Yield Fe = 82.5%
Explanation:
The actual yield is the value produced after an experiment is conducted. The theoretical yield is the value calculated using the balanced chemical equation and atomic/molar masses.
To find the percent yield of iron (Fe), you need to (1) convert grams Al to moles Al (via atomic mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from equation coefficients), then (3) convert moles Fe to grams Fe (via atomic mass), and then (4) calculate the percent yield. It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the sig figs of the given values.
Atomic Mass (Mg): 24.305 g/mol
Atomic Mass (Fe): 55.845 g/mol
3 Mg + 2 FeCl₃ -----> 2 Fe + 3 MgCl₂
20.5 g Mg 1 mole 2 moles Fe 55.845 g
----------------- x ----------------- x ---------------------- x ----------------- =
24.305 g 3 moles Mg 1 mole
= 31.4 g Fe
Actual Yield
Percent Yield = ---------------------------------- x 100%
Theoretical Yield
25.9 g Fe
Percent Yield = -------------------- x 100%
31.4 g Fe
Percent Yield = 82.5%
Answer:
the answer is true
Explanation:
it is the smallest particle in an element that takes part in a chemical reaction
