Answer:14.93
Explanation:heat gained=heat lost
120.5 ×4.184×change=7525
7525÷504.172
Ans14.925
Answer:
the Molecular formula will be; C10H14O
Explanation:
Given the data in the question;
Chrysanthenone is an unsaturated ketone,
it has M+ = 150 and contains 2 double bond(s) and 2 ring(s).
molecular formula = ?
we know that ketone contain 1 oxygen and mass of oxygen is 16
so mass of the C and H remaining will be;
⇒ 150 - 16 = 134
Now we determine the number of C atoms;
⇒ 134 / 13 = 10
hydrocarbon with 10 hydrogen atom have CnH2n+2 means
⇒ ( 10 × 2 ) +2 = 22 hydrogens
But then we have 3 unsaturation meaning 6 hydrogens less and also we have ring meaning 2 more hydrogens
⇒ 22 - 6 - 2 = 14
Hence the Molecular formula will be; C10H14O
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Answer:
The pH of the final solution is 7.15
Explanation:
50 mL of 2.0 M of 
and 25 mL of 2.0 M of 
were mixed to make a solution
Final volume of the solution after dilution = 200 mL
Final concentration of ![K_2HPO_4, [K_2HPO_4] = \frac{50 mL\times 2 M}{200 mL} = 0.5 M](https://tex.z-dn.net/?f=K_2HPO_4%2C%20%5BK_2HPO_4%5D%20%3D%20%5Cfrac%7B50%20mL%5Ctimes%202%20M%7D%7B200%20mL%7D%20%3D%200.5%20M)
Final concentration of![KH_2PO_4, [KH_2PO_4] = \frac{25 mL\times 2 M}{200 mL} = 0.25 M](https://tex.z-dn.net/?f=KH_2PO_4%2C%20%5BKH_2PO_4%5D%20%3D%20%5Cfrac%7B25%20mL%5Ctimes%202%20M%7D%7B200%20mL%7D%20%3D%200.25%20M)
We use Hasselbach- Henderson equation:
![pH = pK_a+ log \frac{[salt]}{[acid]}pka of KH_2PO_4 = 6.85](https://tex.z-dn.net/?f=pH%20%3D%20pK_a%2B%20log%20%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7Dpka%20of%20KH_2PO_4%20%3D%206.85)
Substituting the values:
Therfore, the pH of the final solution is 7.15
Answer : The concentration of 
ion is 0.0375 M.
Explanation :
The balanced equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Pb^{2+}][Cl^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2)
Now put all the given values in this expression, we get:
![2.40\times 10^{-4}=[Pb^{2+}]\times (0.0800)^2](https://tex.z-dn.net/?f=2.40%5Ctimes%2010%5E%7B-4%7D%3D%5BPb%5E%7B2%2B%7D%5D%5Ctimes%20%280.0800%29%5E2)
![[Pb^{2+}]=0.0375M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D0.0375M)
Therefore, the concentration of ion is 0.0375 M.
