Says here the answer is <span>dipole-dipole</span>
Answer:
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
Explanation:

Moles of hydrochloric acid = n
Volume of hydrochloric acid solution = 200.0 mL = 0.200 L
Molarity of the hydrochloric acid = 0.089 M
of HCL

According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.
Then 0.0178 moles of HCl wil be neutralized by :
of sodium bicarbonate
Mass of 0.0178 moles of sodium bicarbonate:
0.0178 mol × 72 g/mol = 1.4952 g
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :
1 mol of urea =15/60.055 = 0.25mol
therefore 200g of water contain 0.25mol
the next step is to determine the malality of our solution in 200g of water, to do this we say:
200 g = 1Kg/1000g = 0.2kg
therefor 0.25mol/0.2Kg = 1.25mol/kg
and from the equation:
we know that i = 1
we are given Kf
b is the molality that we just calculated
therefore;
the solutions freezing point is -2.325°C
Answer:
25%
Explanation:
Ammonium Chloride NH4CI Fertilizer-grade NH4Cl usually contains 25% N. About two-thirds of the world capacity for manufacture of this material is located in Japan, with the remaining one-third situated in India and China.