Both aluminum and chlorine have known charges, which are +3 and -1 respectively. To make them cancel each other out in charge, you would need 3 chlorine and for one aluminum, therefore
would be correct
Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9
Answer:
We can do the nitration of benzene by treating the benzene with a mixture of nitric acid and sulphuric acid by not extending the temperature of 50°C
Explanation:
Nitration of benzene takes place by treating the benzene with a mixture of nitric acid and sulphuric acid at low temperatures such as the temperatures below 50°C
The nitration of benzene takes place through electrophilic substitution reaction
In this reaction the electrophile is nitronium ion (NO2+) which performs an electrophilic substitution reaction on the benzene ring and during the reaction an intermediate will also be formed in which there will be positive charge distributed in the benzene
These electrophile is generated when nitric acid is treated with sulphuric acid
As nitric acid is a strong oxidising agent, here in this case the oxidation state of nitrogen will change from +5 to +3
The reactions regarding the nitration of benzene is present in the file attached
Answer:
0.1
Explanation:
We must first put down the equation of the reaction in order to guide our solution of the question.
2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)
Now from the question, the following were given;
Concentration of acid CA= ??????
Concentration of base CB= 0.299M
Volume of acid VA= 17.8ml
Volume of base VB= 24.7ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
SUBSTITUTING VALUES;
CA= 0.299 × 24.7 ×2 / 17.8×1
CA= 0.8298 M
But;
pH= -log[H^+]
[H^+] = 0.8298 M
pH= -log[0.8298 M]
pH= 0.1
Yo sup??
The answer to this question is option C ie
m/s^2
because the SI unit of distance is m and the SI unit of time is sec
Hope this helps
