In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms
are attached to the C-atom, and one H-atom in the OH group). That means
in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.
The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:
<span>H-atoms in CH3OH = 2 * 6.02 * </span>1023<span> = ~1.2 * 10</span>24
Given:
K = 0.71 = Kp
The reaction of sulphur with oxygen is
S(s) + O2(g) ---> SO2(g)
initial Pressure 6.90 0
Change -x +x
Equilibrium 6.90-x x
Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)
4.899 - 0.71x = x
4.899 = 1.71x
x = 2.86 atm = pressure of SO2 formed
temperature = 950 C = 950 + 273.15 K = 1223.15 K
Volume = 50 L
Let us calculate moles of SO2 formed using ideal gas equation as
PV = nRT
R = gas constant = 0.0821 L atm / mol K
putting other values
n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles
Moles of Sulphur required = 1.42 moles
Mass of sulphur required or consumed = moles X atomic mass of sulphur
mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg of sulphur
I feel like all of them are true
<span>The root mean square velocity of gas molecules as a function of temperature is (M is molecular weight):
v²rms = 3RT/M,
and kinetic energy is given by:
KE = ½Mv ²
We can make the substitution:
KE = ½M3RT/M, or
KE = (3/2)RT.
Notice the molecular weight variable is gone; that means the answer will be the same for F2, Cl2, and Br2:
KE = (3/2) (8.314 J/K*mole) (290 K) = 3617 J/mole.</span>
It is a very large seamount and island created by hot-spot volcanism
