the answer to this question is B
Answer:
molar volume of vapor ethanol is 
mass of ethanol is 15430.22 g
Explanation:
By using ideal gas equation
PV = nRT
Where P is pressure , R is gas constant
so, volume is
molar volume of vapor ethanol is
b)
volume of vessel is given 
therefore total moles of ethanol in given vessel will be
n = 335.44 mol
we know that
mole is given as [/tex]n = \frac{mass}{moleculae weight}[/tex]
weight of ethanol is 46 g/mol
n\times 46 = mass
mass of ethanol is 15430.22 g
Answer:
0.66 moles of NaClO were originally added
Explanation:
When NaClO is added to water, the equilibrium that occurs is:
NaClO(aq) + H₂O(l) ⇄ HClO(aq) + OH⁻(aq)
Where Kb is:
Kb = [HClO] [OH⁻] / [NaClO]
You can obtain Kb from Ka, thus:
Kb = Kw / Ka = 1x10⁻¹⁴ / 3.0x10⁻⁷
Kb = 3.33x10⁻⁸
As pH = 10.50;
pOH = 14 - 10.50 = 3.50
[OH⁻] = 10^{-3.50}
[OH⁻] = 3.16x10⁻⁴M
As OH⁻ and HClO comes from the same equilibrium, [OH⁻] = [HClO]
Replacing in Kb expression:
Kb = [HClO] [OH⁻] / [NaClO]
3.33x10⁻⁸ = [3.16x10⁻⁴] [3.16x10⁻⁴] / [NaClO]
[NaClO] = 0.333M
As there are 2.0L of NaClO solution, moles added were:
2.0L * (0.33moles / L) =
<h3>0.66 moles of NaClO were originally added</h3>
Answer:
Nitrogen, water
Explanation:
"Most common nonmetallic substances such as halogens, halogen acids, sulfur, and phosphorus react with the alkali metals."
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
