I believe that it is liquid.
ANSWER:
London dispersion and hydrogen bonds.
EXPLANATION :
Every molecule experiences London dispersion as an intermolecular force.since the ammonia ion has hydrogen atoms bonded to nitrogen,a very electronegative atom,the molecule is also polar since the nitrogen atom more strongly pulls on the electrons from the hydrogen atoms than the hydrogens themselves do.
This effect is similar to that water,where the oxygen pulls the electrons of the hydrogen atoms with a greater magnitude,resulting in the oxygen having a partial negative charge and the hydrogens having a partial positive charge relative to each other.
This polarity shows that the molecule has dipole-dipole intermolecular forces but since the polarity is from a result of highly electronegative atoms (such as nitrogen,oxygen,fluorine) and hydrogen atoms actually bonded to them,the polarity is categorized in it's own intermolecular force called a hydrogen bond.
I HOPE IT HELPS:)
The balanced equation for the above reaction is as follows;
Na₂SO₄ + BaCl₂ --> BaSO₄ + 2NaCl
Na₂SO₄ reacts with BaCl₂ in the molar ratio 1:1
Number of Na₂SO₄ moles - 10.0 g / 142.1 g/mol = 0.0704 mol
Number of BaCl₂ moles - 10.0 g / 208.2 g/mol = 0.0480 mol
this means that 0.0480 mol of each reactant is used up, BaCl₂ is the limiting reactant and Na₂SO₄ has been provided in excess.
stoichiometry of BaCl₂ to BaSO₄ is 1:1
number of BaSO₄ moles formed - 0.0480 mol
Mass of BaSO₄ - 0.0480 mol x 233.2 g/mol = 11.2 g
theoretical yield is 11.2 g but the actual yield is 12.0 g
the actual product maybe more than the theoretical yield of the product as the measured mass of the actual yield might contain impurities.
percent yield - 12.0 g/ 11.2 g x 100% = 107%
this is due to impurities present in the product or product could be wet.
Answer:
0.57 moles (NH4)3PO4 (2 sig. figs.)
Explanation:
To quote, J.R.
"Note: liquid ammonia (NH3) is actually aqueous ammonium hydroxide (NH4OH) because NH3 + H2O -> NH4OH.
H3PO4(aq) + 3NH4OH(aq) ==> (NH4)3PO4 + 3H2O
Assuming that H3PO4 is not limiting, i.e. it is present in excess
1.7 mol NH4OH x 1 mole (NH4)3PO4/3 moles NH4OH = 0.567 moles = 0.57 moles (NH4)3PO4 (2 sig. figs.)"
