Carbon, helium, and sodium are monoatomic elements.
Hydrogen, iodine, and oxygen are diatomic elements.
[H_{3}O^{+}] = 0.00770 M
The equilibrium equation representing the dissociation of 
Given [H_{3}O^{+}] = 0.00770 M
Let the initial concentration of acid be x and change y
So y = ![[H_{3}O^{+}]](https://tex.z-dn.net/?f=%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%20%20)
=![[FCH_{2}COO^{-}]](https://tex.z-dn.net/?f=%20%5BFCH_%7B2%7DCOO%5E%7B-%7D%5D%20%20)
= 0.00770 M
0.00257 x - 0.00001979 = 0.00005929
x = 0.031 M
Therefore, initial concentration of the weak acid is <u>0.031 M</u>
Answer:
V = 27.98 L
Explanation:
Given data:
Mass of CO₂ = 33.0 g
Pressure = 500 torr
Temperature = 27°C
Volume occupied = ?
Solution:
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 33.0 g/ 44 g/mol
Number of moles = 0.75 mol
Volume of CO₂:
PV = nRT
R = general gas constant = 0.0821 atm.L/ mol.K
Now we will convert the temperature.
27+273 = 300 K
Pressure = 500 /760 = 0.66 atm
By putting values,
0.66 atm×V = 0.75 mol × 0.0821 atm.L/ mol.K × 300 K
V = 18.47 atm.L/0.66 atm
V = 27.98 L
Answer:
Molar concentration of CO₂ in equilibrium is 0.17996M
Explanation:
Based on the reaction:
NiO(s) + CO(g) ⇆ Ni(s) + CO₂(g)
kc is defined as:
kc = [CO₂] / [CO] = 4.0x10³ <em>(1)</em>
As initial concentration of CO is 0.18M, the concentrations in equilibrium are:
[CO] = 0.18000M - x
[CO₂] = x
Replacing in (1):
4.0x10³ = x / (0.18000-x)
720 - 4000x = x
720 = 4001x
x = 0.17996
Thus, concentrations in equilibrium are:
[CO] = 0.18000M - 0.17996 = 4.0x10⁻⁵
[CO₂] = x = <em>0.17996M</em>
<em></em>
Thus, <em>molar concentration of CO₂ in equilibrium is 0.17996M</em>
<em />
I hope it helps!
Poop man poop man poop man poop man
