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disa [49]
2 years ago
15

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
Chemistry
1 answer:
Andrew [12]2 years ago
7 0

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.

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Sergio039 [100]

Answer:

dissolving sugar into water is a physical change

5 0
2 years ago
Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi
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Answer : The value of K_{goal} for the final reaction is, 1.238\times 10^{-7}

Explanation :

The following equilibrium reactions are :

(1) 2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2 K_1=5.40\times 10^{-16}

(2) 2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l) K_2=1.06\times 10^{10}

(3) CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g) K_3=2.68\times 10^{-9}

The final equilibrium reaction is :

CO_2(g)\rightleftharpoons C(s)+O_2(g) K_{goal}=?

Now we have to calculate the value of K_{goal} for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

K_{goal}=\sqrt{K_1\times K_2\times K_3}

Now put all the given values in this expression, we get :

K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}

K_{goal}=1.238\times 10^{-7}

Therefore, the value of K_{goal} for the final reaction is, 1.238\times 10^{-7}

3 0
3 years ago
A mixture of oxygen, hydrogen, and nitrogen exerts a total pressure of 378 kPa. If the partial pressures of oxygen and hydrogen
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What's the relationship between total and partial pressure? The total pressure is the sum of the parcial pressures!


So for us, it would be:

378= 212+101+x

where x is the parcial pressure of nitrogen.

Now we count:
378= 212+101+x
378=313+x
378-313=x
65=x

So the parcial pressure exerted by nitrogen is 65!

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Answer: Jose and Sarah are running in the cross country meet. Their skeletal and muscular systems work to move them through the course. As they run, they breathe faster and harder and their heart rate increases.

A.work together

Explanation:This should help:)

5 0
3 years ago
What is the concentration of a solution containing 0.32 moles of NaCl in 3.4 liters?
dsp73
Get the molarity we need to divide the number of moles of NaCl by the volume of the solution. So, 0.32 moles NaCl divided by 3.4 L, and that gives 0.094 M NaCl.
5 0
2 years ago
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