1.turn the revolving turret so that low power lens come into position
2 place the microscope slide on the stage
3 look at the objective lens and the stage from the side as you turn adjust knob so that stage can move upward
4 look through the eyepiece and move the focus knob until image come into focus
5.adjust the condenser to regulate thee amount of light
6 move the microscope slide around until the sample is in the center of field of view
Answer:
This is a pretty straightforward example of how an ideal gas law problem looks like.
Your strategy here will be to use the ideal gas law to find the pressure of the gas, but not before making sure that the units given to you match those used by the universal gas constant.
So, the ideal gas law equation looks like this
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
P
V
=
n
R
T
a
a
∣
∣
−−−−−−−−−−−−−−−
Here you have
P
- the pressure of the gas
V
- the volume it occupies
n
- the number of moles of gas
R
- the universal gas constant, usually given as
0.0821
atm
⋅
L
mol
⋅
K
T
- the absolute temperature of the gas
Take a look at the units given to you for the volume and temperature of the gas and compare them with the ones used in the expression of
R
.
a
a
a
a
a
a
a
a
a
a
a
Need
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
Have
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
Liters, L
a
a
a
a
a
a
a
a
a
a
a
a
a
Liters, L
a
a
a
a
a
a
a
a
a
a
a
√
a
a
a
a
a
a
a
Kelvin, K
a
a
a
a
a
a
a
a
a
a
a
a
Celsius,
∘
C
a
a
a
a
a
a
a
a
a
×
Notice that the temperature of the gas must be expressed in Kelvin in order to work, so make sure that you convert it before plugging it into the ideal gas law equation
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
T
[
K
]
=
t
[
∘
C
]
+
273.15
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
Rearrange the ideal gas law equation to solve for
P
P
V
=
n
R
T
⇒
P
=
n
R
T
V
Plug in your values to find
P
=
0.325
moles
⋅
0.0821
atm
⋅
L
mol
⋅
K
⋅
(
35
+
273.15
)
K
4.08
L
P
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
2.0 atm
a
a
∣
∣
−−−−−−−−−−−
The answer is rounded to two sig figs, the number of sig figs you have for the temperature of the gas.
Answer:
t = 3,496x10⁹ years
Explanation:
The decay of ²³⁸U is:
²³⁸U → ²⁰⁶Pb + 8He + 6e⁻
Moles of ²⁰⁶Pb presents in 0,623mg are:
0,623x10⁻³g×(1mol / 206g) = 3,02x10⁻⁶ moles of ²⁰⁶Pb.
These moles are equals to moles of ²³⁸U before decay, that means, 3,02x10⁻⁶ moles²³⁸U
In grams:
3,02x10⁻⁶ moles²³⁸U× (238g / 1mol) = 7,20x10⁻⁴ g ²³⁸U = 0,720 mg²³⁸U
That means initial ²³⁸U was 1,000mg + 0,720mg =<em> 1,720mg</em>
Applying the formula:
ln (N₀/N) t₁₂ = t ln2
Where N₀ is initial amount of uranium (1,720mg), N is concentration of uranium (1,000mg), half-life time is a constant (t₁₂= 4,468x10⁹ years) and t is the time transcurred for the reaction. Replacing:
ln(1,720/1)*4,468x10⁹ years = t ln2
<em>t = 3,496x10⁹ years</em>
<em></em>
I hope it helps!
Answer:
The lowest region of the atmosphere
