Answer:
a. 5.77L
b. 700mmHg
c. 395K
Explanation:
Using PV = nRT we can solve these problems where:
P is pressure of the gas in atm (1atm = 760mmHg)
V is volume in liters
n are moles of the gas
R is gas constant: 0.082atmL/molK
T is asbolute temperature in K
a. PV = nRT
V = nRT/P
P = 773mmHg*(1atm/760mmHg) = 1.017atm
T = 25°C+273 = 298K
V = 0.240mol*0.082atmL/molK*298K / 1.017atm
V = 5.77L
b. PV = nRT
P = nRT/V
P = 0.0947mol*0.082atmL/molK*309K/0.635L
P = 0.9216atm * (760mmHg/1atm) = 700mmHg
c. PV = nRT
PV/nR = T
P = 727mmHg * (1atm / 760mmHg) = 0.9566atm
0.9566atm*13.3L/0.393mol*0.082atmL/molK = T
T = 395K
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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Your question looks a bit incomplete as you have the same contents in options a) and d). According to your list, I can't see the correct answer, but I can give you one.The difference between the potential energy of the products of the potential energy of the reactants is equal to the enthalpy of the reaction.
Answer:
.
Explanation:
Electrons are conserved in a chemical equation.
The superscript of 
indicates that each of these ions carries a charge of 
. That corresponds to the shortage of one electron for each 
ion.
Similarly, the superscript 
on each 
ion indicates a shortage of three electrons per such ion.
Assume that the coefficient of (among the reactants) is 
, and that the coefficient of (among the reactants) is 
.
.
There would thus be silver (
) atoms and aluminum (
) atoms on either side of the equation. Hence, the coefficient for 
and 
would be 
and 
, respectively.
.
The ions on the left-hand side of the equation would correspond to the shortage of electrons. On the other hand, the 
ions on the right-hand side of this equation would correspond to the shortage of 
electrons.
Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of electrons, the right-hand side should also be electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of electrons. These two expressions should have the same value. Therefore, 
.
The smallest integer and that could satisfy this relation are 
and 
. The equation becomes:
.
