Answer:
See explanation
Explanation:
Sr(s) + 2HCl(aq) -----> SrCl2(aq) + H2(g)
Ionically;
Sr(s) + 2Cl^-(aq) ----> SrCl2(aq)
If we look at the reaction above, strontium atom was dissolved in hydrochloric acid. The strontium atom is now oxidized by the acid to give Sr^2+ ion according to the equation shown above.
Answer:
pH=10.97
Explanation:
the solution of methyl amine with methylammonium chloride will make a buffer solution.
The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:
![pOH=pKb+log\frac{[salt]}{[base]}](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D)
pH = 14- pOH
Let us calculate pOH

[Salt] = [methylammonium chloride] = 0.10 M (initial)
After adding base
![[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M](https://tex.z-dn.net/?f=%5Bsalt%5D%20%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X20%7D%7B%2820%2B50%29%7D%3D%200.0286M)
[base] = [Methylamine]=0.10
After mixing with salt
![[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M](https://tex.z-dn.net/?f=%5Bbase%5D%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X50%7D%7B%2820%2B50%29%7D%3D%200.0714M)
pKb= -log[Kb]= 3.43
Putting values
pOH = ![3.43+log(\frac{[0.0286]}{0.0714}](https://tex.z-dn.net/?f=3.43%2Blog%28%5Cfrac%7B%5B0.0286%5D%7D%7B0.0714%7D)
Tetraphosphorus Decaoxide has a molecular formula of P4O10,
all else are known. Therefore the balanced chemical equation would be:
4PH3 + 8O2 --> 6H2O + P4O10
Taking into account the phases:
4PH3(g) + 8O2(g) --> 6H2O(g) + P4O10(s)
The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Each mole of propane requires 5 moles of oxygen.
Answer:
3.43 %
Explanation:
We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.
0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2
0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce
.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce
0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %