80.0 liters of oxygen gas is collected over water at 50 degrees celcius. the atomsphereic pressure in the room is 96.00 kPA. wha
1 answer:
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Answer:
The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml
Explanation:
Here we have the reaction of AgNO₃ and NaCl as follows;
AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)
Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,
Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains
0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃
Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;
0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL
Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.
Answer:
is the solubility of nitrogen gas in a diver's blood.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
where,
= Henry's constant =
= partial pressure of nitrogen
(Raoult's law)
is the solubility of nitrogen gas in a diver's blood.