Answer:
Q = mcT ...you can either substitute the molar heat capacity of water in the place of c or the specific heat capacity of water.
Explanation:
Answer:
Percent yield = 50%
Explanation:
Given data:
Mass of CH₄ = 16 g
Mass of O₂ = 32 g
Mass of CO₂ = 11 g
Percent yield of CO₂ = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass/ molar mass
Number of moles = 16 g /16 g/mol
Number of moles = 1 mol
Number of moles of O₂:
Number of moles = mass/ molar mass
Number of moles = 32 g /32 g/mol
Number of moles = 1 mol
Now we will compare the moles of CO₂ with both reactant.
O₂ : CO₂
2 : 1
1 : 1/2×1= 0.5 mol
CH₄ : CO₂
1 : 1
Number of moles of CO₂ produced by oxygen are less so it will limiting reactant.
Theoretical yield:
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 0.5 mol × 44 g/mol
Mass = 22 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 11 g/ 22 g × 100
Percent yield = 50%
Answer:
The
expression for the weak base equilibrium is:
![K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
Explanation:

The expression of the equilibrium constant of base
can be given as:
![K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%5BH_2O%5D%7D)
]![K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%20%5BH_2O%5D%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
As we know, water is pure solvent, we can put ![[H_2O]=1](https://tex.z-dn.net/?f=%5BH_2O%5D%3D1)
![K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%201%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
So, the the
expression for the weak base equilibrium is:
![K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
Answer:
285g of fluorine
Explanation:
To solve this problem we need to find the mass of Freon in grams. Then, with its molar mass we can find moles of freon and, as 1 mole of Freon, CCl₂F₂, contains 2 moles of fluorine, we can find moles of fluorine and its mass:
<em>Mass Freon:</em>
<em>2.00lbs * (454g / 1lb) = </em>908g of Freon
<em>Moles freon -Molar mass: 120.91g/mol- and moles of fluorine:</em>
908g of Freon * (1mol / 120.91g) =
7.5 moles of freon * (2moles Fluorine / mole Freon): 15 moles of fluorine
<em>Mass fluorine -Atomic mass: 19g/mol-:</em>
15 moles F * (19g / mol) =
<h3>285g of fluorine</h3>
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