Answer:
Explanation:
Firstly of all we have to construct the min-heap of the k-sub list and each sub list which is a node in the constructed min-heap.
We have several steps to follows:
Step-1. When we compare the two sub lists, at the starting we can compare their first elements which is actually their minimum elements.
Step-2. The min-heap formation will cost be O(k) time.
Step-3. After the step 1 & step 2 we can run the minimum algorithm which can be extracted from the minimum element in the root list.
Step-4. Then Update the root list in the heap and after that simplify the min-heap as maintained by the new minimum element in the root list.
Step-5. If any root sub-list becomes empty in the step 4 then we can take any leaf sub-list from the root and simplify it.
Step-6. At every Extraction of the element it can take up to O(log k) time.
Hence, We can say that the extract of n element in the total whose
Running time will be O(n log k + k) which can be equal to the O(n log k+ k) (since k < n).
At the equivalence point, the NaF completely reacts with the HCl
The correct option that gives the equivalence point when titrating NaF solution with HCl(aq), occur at a pH value is option c);
c) <u>Below 7 because it is determined by HF (aq)</u>
Reason:
The ionic equation of the reaction taking place in the titration is presented as follows;
- Na⁺ (aq) + F⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → HF (aq)+ Na⁺(aq) + Cl⁻(aq)
Therefore, by removing the spectator ions, we get;
Therefore;
The net ionic equation is the formation of hydrofluoric acid, HF, which is a
weak acid, therefore, the pH value will be <u>below 7 because it is determined </u>
<u>by HF</u>
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Learn more about equivalence point here: titration
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Answer:
228.7
Explanation:
According to Boyle's law
P1V1 = P2V2
where P1& P2 are initial and final pressure
and V1& V2 and initial and final volumes respectively.
Given:
P1 = 136atm, P2 = 131.6atm
V1 = 142; V2 = ?
V2 = 
= 
= 146.7
Total volume, V = V1 + V2
=146.7 + 142
<u>= 228.7</u><u></u>
