4 weeks for the moon to go to the next full moon
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Explanation:
It is known that
value of acetic acid is 4.74. And, relation between pH and
is as follows.
pH = pK_{a} + log ![\frac{[CH_{3}COOH]}{[CH_{3}COONa]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOH%5D%7D%7B%5BCH_%7B3%7DCOONa%5D%7D)
= 4.74 + log 
So, number of moles of NaOH = Volume × Molarity
= 71.0 ml × 0.760 M
= 0.05396 mol
Also, moles of
= moles of 
= Molarity × Volume
= 1.00 M × 1.00 L
= 1.00 mol
Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

Initial : 1.00 mol 1.00 mol
NaoH addition: 0.05396 mol
Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)
= 0.94604 mol = 1.05396 mol
As, pH = pK_{a} + log ![\frac{[CH_{3}COONa]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOONa%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
= 4.74 + log 
= 4.69
Therefore, change in pH will be calculated as follows.
pH = 4.74 - 4.69
= 0.05
Thus, we can conclude that change in pH of the given solution is 0.05.
Answer:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Explanation:
Hello,
In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

The suitable equilibrium constant turns out:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Or in terms of the initial equilibrium constant:

Since the second reaction is a doubled version of the first one.
Best regards.
Answer:
0.84 moles of oxygen are required.
Explanation:
Given data:
Mass of CO₂ produced = 37.15 g
Number of moles of oxygen = ?
Solution:
Chemical equation:
C + O₂ → CO₂
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 37.15 g/ 44 g/mol
Number of moles = 0.84 mol
Now we will compare the moles of oxygen and carbon dioxide.
CO₂ : O₂
1 : 1
0.84 : 0.84
0.84 moles of oxygen are required.