Answer:
The final balanced equation is
Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+
Explanation:
It is given that sodium hydroxide is added to collect the solid nickel(II) hydroxide product
The empirical equation for this statement is
Ni2+ + NaOH --> Ni (OH)2 + Na+
We will first balance the hydroxide molecule. On the right side there are two OH molecules.
Thus, on the left side we will take 2 sodium hydroxide
Ni2+ + 2NaOH --> Ni (OH)2 + Na+
Now we will balance the sodium ion which are 2 in numbers on the left side and 1 on the right side
Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+
So, the final balanced equation is
Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+
Answer:
the number of carbon and hydrogen atoms present in the molecule of Ethane that is it contains two carbon atoms and six hydrogen atoms
Explanation:
Answer:
a)calculated molarity of NaOH would be lower
b) calculated molarity of NaOH would be lower
c) calculated molarity of NaOH would be lower
d) calculated molarity of NaOH would be unaffected
Explanation:
Let us recall that the reaction of NaOH and HCl is as follows;
NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)
Since the reaction is 1:1, when the number of moles of HCl reacting with NaOH is low due to dilution, the calculated molarity of NaOH also becomes less than it's accurate value.
When 40mL of water is added to the titration flask rather than 25ml of water, the acid is more dilute hence less number of moles of acid than necessary reacts with the base thereby yielding a less than accurate value of the molarity of NaOH.
If the burette wet with water is not rinsed with NaOH solution, the concentration of the NaOH in the burette decreases due to dilution with water and a less than accuracy value is calculated for the molarity of NaOH.
If five drops of phenolphthalein is used instead of one or two drops, there is no qualms since enough phenolphthalein may be added to ensure that a sharp end point is obtained.
<u>Answer:</u> The freezing point of solution is 2.6°C
<u>Explanation:</u>
To calculate the depression in freezing point, we use the equation:
Or,
where,
= 
Freezing point of pure solution = 5.5°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal freezing point depression constant = 5.12 K/m = 5.12 °C/m
= Given mass of solute (anthracene) = 7.99 g
= Molar mass of solute (anthracene) = 178.23 g/mol
= Mass of solvent (benzene) = 79 g
Putting values in above equation, we get:
Hence, the freezing point of solution is 2.6°C
