Answer:
This is known as the coefficient factor
Explanation:The balanced equation makes it possible to convert information about one reactant or product to quantitative data about another element.
Answer:
20 g/mol
Explanation:
We can use <em>Graham’s Law of diffusion</em>:
The rate of diffusion (<em>r</em>) of a gas is inversely proportional to the square root of its molar mass (<em>M</em>).
If you have two gases, the ratio of their rates of diffusion is
Squaring both sides, we get
Solve for <em>M</em>₂:
a. 30 moles of H₂O
b. 2.33 moles of N₂
<h3>Further explanation</h3>
Given
a. 20 moles of NH₃
b. 3.5 moles of O₂
Required
a. moles of H₂O
b. moles of N₂
Solution
Reaction
4NH₃+3O₂⇒2N₂+6H₂O
a. From the equation, mol ratio NH₃ : H₂O = 4 : 6, so mol H₂O :
=6/4 x mol NH₃
= 6/4 x 20 moles
= 30 moles
b. From the equation, mol ratio N₂ : O₂ = 2 : 3, so mol N₂ :
=2/3 x mol O₂
= 2/3 x 3.5 moles
= 2.33 moles
Answer:
yes
Explanation:
Solubility is an observation and no chemical reaction takes place. The composition of the compound/element is not changed.
- Hope that helped! Please let me know if you need further explanation.
Answer:
MM = 680g / mol
Explanation:
Hello! To calculate the molar mass of the compound (g / mol), I first have to calculate the molarity.
Molarity can be calculated from the osmotic pressure equation.
op = M * R * T
op = osmotic pressure = 28.1mmHg * (1 atm / 760mmHg) = 0.037atm
M = molarity
R = gas constant
T = temperature (K) = 20 ° C + 273.15 = 293.15K
M (mol / L) = op / R * T
M = 0.037atm / ((0.082 (atm * L) / (K * mol)) * 293.15K) = 0.0015mol / L
As I have the volume = 100ml * (1L / 1000ml) = 0.1L
I can calculate the amount of moles
n = M * V = 0.0015 * 0.1 = 0.00015mol
n = m / MM
m = mass
MM = molar mass
MM = m / n = 0.102g / 0.00015mol
MM = 680g / mol
