for 39g water solute dissolved at 20C = solubility ( g/ 100 g H2O ) × mass of water = ( 11g / 100g H2O ) × 39g H2O = 4.29 g
amount of solute dissolved at 30 C =
= 23 / 100 * 39 = 8.97 g
Amount of extra solute dissolved = 8.97 - 4.29 = 4.7 g
Answer:
Physical Properties
Explanation:
You are able to see physical properties but are unable to see chemical properties.
This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:
pH = pKa + log[(salt/acid]
Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).
Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L
Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M
Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
Answer:
Don't post any question if isn't related to the topic or to your homework or assignment.
Explanation:
I’m assuming you mean barium nitrite, Ba(NO2)2.
First convert grams of Ba(NO2)2 to moles using the molar mass of Ba(NO2)2. Then use the mole ratio of 4 moles of oxygen per 1 mole of Ba(NO2)2 to convert to moles of oxygen. Then use the molar mass of oxygen to convert to grams of oxygen.
45.7 g Ba(NO2)2 • 1 mol Ba(NO2)2 / 229.35 g Ba(NO2)2 • 4 mol O / 1 mol Ba(NO2)2 • 16.0 g O / 1 mol O = 12.8 g oxygen
