the answer is in the picture, btw the molar mass for the first one is wrong, it should be 77.98, and the final product is 2.32
Answer:
69.7% is percent yield
Explanation:
Based on the reaction:
3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)
2 moles of Na3PO4 react producing 6 moles of NaNO3.
As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:
24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =
72.6 moles of NaNO3
As there are produced 50.6 moles of NaNO3, percent yield is:
50.6 moles NaNO3 / 72.6 moles NaNO3 =
<h3>69.7% is percent yield</h3>
7.5 M is the concentration of 60 ml of H3PO4 if it is neutralized by 225 ml of 2 M Ba(OH)2.
Explanation:
Data given:
volume of phosphoric acid, Vacid =60 ml
volume of barium hydroxide, Vbase = 225 ml
molarity of barium hydroxide, Mbase = 2M
Molarity of phosphoric acid, Macid =?
the formula for titration is used as:
Macid x Vacid = Mbase x Vbase
rearranging the equation to get Macid
Macid = 
Macid =
Macid = 7.5 M
the concentration of the phosphoric acid is 7.5 M and the volume is 60 ml. Thus 7.5 M solution of phosphoric acid is used to neutralize the barium hydroxide solution of 2M.
Answer:
We know that
ħf = ф + Ekmax
where
ħ = planks constant = 6.626x10^-34 J s
f = frequency of incident light = 1.3x10^15 /s (1 Hz =
1/s)
ф = work function of the cesium = 2.14 eV
Ekmax = max kinetic energy of the emmitted electron.
We distinguish that:
1 eV = 1.602x10^-19 J
So:
2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J
So,
Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J
= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J
Answer:
5.19x10^-19 J
Kinetic energy:
In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.
Answer details:
Subject: Chemistry
Level: College
Keywords:
• Energy
• Kinetic energy
• Kinetic energy of emitted electrons
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