The given question is incomplete. The complete question is:
The combustion of propane (C3H8) in the presence of excess oxygen yields
and
When only 2.5 mol of
are consumed in order to complete the reaction, ________ mol of
are produced.
Answer: Thus when 2.5 mol of
are consumed in their reaction, 1.5 mol of
are produced
Explanation:
The balanced chemical equation is:

According to stoichiometry :
5 moles of
produce = 3 moles of 
Thus 2.5 moles of
will produce =
moles of 
Thus when 2.5 mol of
are consumed in their reaction, 1.5 mol of
are produced
Answer:
chemistry - the science that studies the properties of substances and natural fenomens .
...........
Answer:
136.63 °C
Explanation:
ΔTb=Tb solution - Tb pure
Where; Tb pure = 133.60°C
molar mass of solute = 121.14 g/mol
number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles
molality = 0.431 moles/350 * 10^-3 = 1.23 molal
Then;
ΔTb = Kb * m * i
Kb = 2.46°C kg mol^-1
m = 1.23 molal
i = 1
ΔTb = 2.46 * 1.23 * 1
ΔTb = 3.03 °C
Hence;
Tb solution = ΔTb + Tb pure
Tb solution = 3.03 °C + 133.60°C
Tb solution = 136.63 °C
The idea is that all continents move at a rate of approximately 2 inches per year because their moving ever since pangea (all continents were one) separated
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:

Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)


Therefore, the pH of the solution is 11.48.
I hope it helps you!