<h3>Answer:</h3>
0.64 Moles of Propane
<h3>Explanation:</h3>
Data:
Moles of Carbon = 1.5 mol
Conversion factor = 7 mol C produces = 3 mol of Propane
Solution:
As we know,
7 moles of Carbon produces = 3 moles of Propane
Then,
1.5 moles of Carbon will produce = X moles of Propane
Solving for X,
X = (1.5 moles × 3 moles) ÷ 7 moles
X = 0.6428571 moles of Propane
Or rounded to two significant figures,
X = 0.64 Moles of Propane
Wait hold up do you live in Wilson county????
<u>Given:</u>
H2(g) + Cl2 (g) → 2HCl (g)
<u>To determine:</u>
The enthalpy of the reaction and whether it is endo or exothermic
<u>Explanation:</u>
Enthalpy of a reaction is given by the difference between the enthalpy of formation of reactants and products
ΔH = ∑nHf (products) - ∑nHf (reactants)
= [2Hf(HCl)] - [Hf(H2) + Hf(Cl2)] = 2 (-92.3) kJ = - 184.6 kJ
Since the reaction enthalpy is negative, the reaction is exothermic
<u>Ans:</u> The enthalpy of reaction is -184. kJ and the reaction is exothermic
Answer:
![\frac{[F^{-}]}{[HF]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BF%5E%7B-%7D%5D%7D%7B%5BHF%5D%7D)
is larger
Explanation:
, where 
is the acid dissociation constant.
For a monoprotic acid e.g. HA, ![K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
and ![\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
So, clearly, higher the value , lower will the the 
In this mixture, at equilibrium, ![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
will be constant.
of HF is grater than of HCN
Hence, ![(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})](https://tex.z-dn.net/?f=%28%5Cfrac%7BF%5E%7B-%7D%7D%7B%5BHF%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HF%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29%3E%28%5Cfrac%7BCN%5E%7B-%7D%7D%7B%5BHCN%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HCN%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29)
So, is larger
Answer:
6.022 × 10²² atoms
Explanation:
Generally 1 mol of any element contains 6.02×10^23 atoms. The number 6.022 × 10²³ is known as Avogadro's number.
Mass of Aluminium = 2.70g
Molar mass = 27g/mol
Number of moles = Mass / Molar mass = 2.70 / 27 = 0.1 mol
1 mol = 6.022 × 10²³
0.1 mol = x
x = 6.022 × 10²³ * 0.1 = 6.022 × 10²² atoms
