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
Answer:
bombarding it with an energetic particle
Explanation: nuclear reaction, a change in the identity or characteristics of an atomic nucleus, induced by bombarding it with an energetic particle. The bombarding particle may be an alpha particle, a gamma-ray photon, a neutron, a proton, or a heavy-ion.
Answer:
Steel and cast iron
Explanation:
They are all metal but assuming that you are finding the best material for your pan i suggest going for steel or cast iron
Answer: 104 g
Explanation: reaction Cr2O3 + 3 H2 ⇒ 2 Cr + 3 H2O
M(Cr2O3) = 150 g/mol, so n = m/M = 1.0 mol
Number of moles of H2 should be 3.0 moles and
It is much greater (150 g / 2.016 g/mol)
1 mol Cr2O3 produces 2 mol Cr.
Mass m= 2.0 mol· 52g/mol= 104 g
Answer:
0.007 mol
Explanation:
We can solve this problem using the ideal gas law:
PV = nRT
where P is the total pressure, V is the volume, R the gas constant, T is the temperature and n is the number of moles we are seeking.
Keep in mind that when we collect a gas over water we have to correct for the vapor pressure of water at the temperature in the experiment.
Ptotal = PH₂O + PO₂ ⇒ PO₂ = Ptotal - PH₂O
Since R constant has unit of Latm/Kmol we have to convert to the proper unit the volume and temperature.
P H₂O = 23.8 mmHg x 1 atm/760 mmHg = 0.031 atm
V = 1750 mL x 1 L/ 1000 mL = 0.175 L
T = (25 + 273) K = 298 K
PO₂ = 1 atm - 0.031 atm = 0.969 atm
n = PV/RT = 0.969 atm x 0.1750 L / (0.08205 Latm/Kmol x 298 K)
n = 0.007 mol
