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
The primary reason for this was that Mendeleev
didn't know that atomic numbers actually existed. Atomic numbers
were only discovered a period after Mendeleev's time. The use of X-rays made it
possible to find the atomic number, and those had not been discovered yet. <span>
<span>The periodic table was then arranged in 1913 by Henry Moseley
in an arrangement according to atomic number.</span></span>
that's your work bro do by your own
Answer:
The partial pressure of argon in the flask = 71.326 K pa
Explanation:
Volume off the flask = 0.001 
Mass of the gas = 1.15 gm = 0.00115 kg
Temperature = 25 ° c = 298 K
Gas constant for Argon R = 208.13 
From ideal gas equation P V = m RT
⇒ P = 
Put all the values in above formula we get
⇒ P = 
× 208.13 × 298
⇒ P = 71.326 K pa
Therefore, the partial pressure of argon in the flask = 71.326 K pa
