For every meter, the equivalent measurements is 1000 millimeters. Hence in the problem where the number of millimeters is given, we divide the number by 1000 to get the number of meters. The answer here is 0.01123 m.
Explanation:
Z = atomic mass of the element and , A = atomic mass of the element .
a) Z = 11, A = 23
Element = Sodium
symbol: ²³₁₁Na .
b) Z = 28, A = 64
Element = Nickel
symbol: ⁶⁴₂₈Ni .
c) Z = 50, A = 115
Element = tin
symbol: ¹¹⁵₅₀Sn .
d) Z = 20, A = 42
Element = Calcium
symbol: ⁴²₂₀Ca .
Answer:
PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.
PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.
PART C: Dipole dipole and in some cases hydrogen bonding operate between the hydrogen atom of a polar bond and a nearby small electronegative atom. Only if the atom bonded to it were F, O or N it would be hydrogen bonding. Otherwise it is dipole dipole.
These problems are a bit interesting. :)
First let's write the molecular formula for ammonium carbonate.
NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)
17.6 gNH4CO3
Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.
17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)
Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)
NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol
17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4
Now just take the molar mass we found to convert that amount into moles!
4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4
