N: 60 mol
p: 20 atm = 20×1013hPa = 20260 hPa
T: 50°C = 323 K
R: 83,14 hPa·dm³/mol·K
V: ?
.................
pV = nRT
V = nRT/p
V = (60×83,14×323)/20260
V = 79,529 dm³ ≈ 79,6 dm³
79,6 dm³ = 79,6 L
:•)
Answer:
45.2%
Explanation:
To calculate the percent of an element in a compound we divide the molar mass of the element by the compound and multiply that by 100
First lets find the molar mass of Fluoride
Looking at the periodic table Fluoride has a molecular mass of 18.998 g
Now we need to find the molecular mass of NaF
Looking at a periodic table, Sodium (Na) has a molecular mass of 22.990g and Fluoride has a molecular mass of 18.998 so NaF has a molecular mass of 22.990(1) + 18.998(1) = 41.988g
Now we divide the mass of fluoride by the mass of sodium fluoride and multiply that by 100 to find the percentage of fluoride that is present in NaF
Mass of Fluoride = 18.998g
Mass of Sodium Fluoride = 41.988g
Percentage of fluoride present in NaF = (18.998g / 41.988g) * 100 = 45.2%
London dispersion
dipole-dipole
<h3><u>Answer;</u></h3>
True
<h3><u>Explanation</u>;</h3>
- The molecule NH3 contains all single bonds.
- NH3 has a three single covalent bond among its nitrogen and hydrogen atoms,because one valence electron of each of three atom of hydrogen is shared with three electron.
- There are three covalent bonds are in NH3 . Each hydrogen make a single bond with nitrogen and there is also a pair of electron which is unpaired from nitrogen.
