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hjlf
3 years ago
9

Why does the carbon anode burn away in the electrolysis of aluminium chloride?

Chemistry
2 answers:
Xelga [282]3 years ago
5 0

Answer:

umm nice questionhdhssgsishsisbsiegeiehei

almond37 [142]3 years ago
3 0
Carbon dioxide is formed when oxygen reacts with carbon in positive electrode which eventually burns out. This is why the carbon anode burn away in the electrolysis of aluminium chloride.
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A sample of magnesium is burned in oxygen to form magnesium oxide. What mass of oxygen is consumed if 74.62 g magnesium oxide is
lana [24]
74.62 g of magnesium oxide is formed from 45.00 g magnesium so 74.62-45.00= 29.62 g of oxygen is consumed or in other words a new compound is formed in the burning of magnesium in oxygen with a heavier mass than the pure magnesium.
3 0
3 years ago
5.25 ml of substance A has a mass of 3.9 g and 6.24 ml of substance B has a mass of 4.4 g. Which liquid is more dense?
Viktor [21]
Substance A because it weighs less
5 0
3 years ago
A scientist carries out a reaction twice. In Trial A, the scientist combines the reactants in a container. In Trial B, the scien
Paha777 [63]
It would be C because It will have a lower activation energy than Trial A.
7 0
4 years ago
Read 2 more answers
A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentratio
Pachacha [2.7K]

Answer:

pK_{a} of HA is 6.80

Explanation:

pK_{a}=-logK_{a}

Acid dissociation constant (K_{a}) of HA is represented as-

                K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}

So, K_{a}=1.6\times 10^{-7}

Hence pK_{a}=-log(1.6\times 10^{-7})=6.80

6 0
3 years ago
The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?
sweet [91]

Answer : The maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

Explanation :

The solubility equilibrium reaction will be:

                       Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-

Initial conc.                        0.220       0

At eqm.                             (0.220+s)   2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ni^{2+}][CN^-]^2

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

3.0\times 10^{-23}=(0.220+s)\times (2s)^2

s=5.84\times 10^{-12}M

Therefore, the maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

7 0
3 years ago
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