K:
m=155g
M=39g/mol
n = 155g / 39g/mol ≈ 3,97mol
KNO₃:
m=122g
M=101g/mol
n = 122g/101g/mol = 1,21mol
2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent
KNO₃ is limiting reagent
CH3 is the empirical formula for the compound.
A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.
The number of atom or moles in the compound is
1.17 g C X 1 mol of C / 12.011 g C = 0.097411 mol of C.
0.287 g H x 1 mol of H / 1 g H = 0.28474 mol H.
This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.
So we can represent the compound with the formula C0.974H0.284.
Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.
we can divide 0.284 by 0.0974.
0.284 / 0.0974 = 3.
So here, Carbon is one and hydrogen is 3.
We can write the above formula as a CH3.
Hence the empirical formula for the sample compound is CH3.
For a detailed study of the empirical formula refer given link brainly.com/question/13058832.
#SPJ1.
The standard ambient temperature and pressure are
Temperature =298 K
Pressure = 1atm
The density of gas is 1.5328 g/L
density = mass of gas per unit volume
the ideal gas equation is
PV = nRT
P = pressure = 1 atm
V = volume
n = moles
R= gas constant = 0.0821 Latm/mol K
T = 298 K
moles = mass / molar mass
so we can write
n/V = density / molar mass
Putting values
Thus molar mass of gas is 37.50g/mol
Answer:
i am new but I haven't reached this level yet
Nitrogen can make bonds with other atoms.. Typically though it only makes 3 bonds, so it fills its octet.
