A 1.00 L flask is filled with 1.30 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres
1 answer:
Answer :
Part A : The partial pressure of argon is, 0.795 atm
Part B : The partial pressure of ethane is, 0.505 atm
Explanation :
<u>Part A :</u>
First we have to calculate the moles of argon.
Molar mass of argon = 39.95 g/mole
Now we have to calculate the partial pressure of argon.
Using ideal gas equation :
PV = nRT
where,
P = Pressure of argon = ?
V = Volume of argon = 1.00 L
n = number of moles of argon = 0.0325 mole
R = Gas constant =
T = Temperature of argon =
Putting values in above equation, we get:
The partial pressure of argon is, 0.795 atm
<u>Part B :</u>
Now we have to calculate the partial pressure of ethane.
As we know that,
Total pressure = Partial pressure of argon + Partial pressure of ethane
1.300 atm = 0.795 atm + Partial pressure of ethane
Partial pressure of ethane = 1.300 - 0.795
Partial pressure of ethane = 0.505 atm
The partial pressure of ethane is, 0.505 atm
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Answer:
The answer to the question is;
The required torque that it would take to cause the gyroscopes to precess through an angle of 1.0×10−6 degree during a 5.0-hour exposure of a galaxy is 2.44 ×10⁻¹² N·m
Explanation:
To solve the question we first resolve the units of the given quantities as follows
The gyroscopes spin at 19,200 rpm that is 19,200 revolutions per minute
1 revolution = 2π rad and
1 minute = 60 seconds
Therefore 19,200 revolutions per minute = 2π×19,200÷60 rad/s
= 2010.619 rad/s
The angle of precess is given as 1.0×10⁻⁶ °. We convert the angle to radians as follows
360 ° = 2π radians
1 ° = radians and
1.0×10⁻⁶ ° = radians × 1.0×10⁻⁶ ° = 1.745×10⁻⁸ rad
To find the torque we note that the torque is given by
Precession angular speed × The moment of inertia × angular velocity
The precession angular speed is given by
The precession angle was determined in rad as 1.745×10⁻⁸ rad
The precession time is 5 hours which is equal to 5×60×60 = 18000 s
Therefore the precession velocity = = 9.696×10⁻¹³ rad/s
The moment of inertia is given by
Formula for the moment of inertia of a thin walled cylinder I = m·r²
Where:
r = Radius of the gyroscope = Diameter/2 = 5.0 cm/2 = 2.5 cm = 0.025 m
m = Mass of each gyroscopes = 2.0 kg
Therefore I = m·r² = 2.0 kg × (0.025 m)² = 0.00125 kg·m²
Torque, τ = Ω·I·ω
Where:
Ω = Precession velocity = 9.696×10⁻¹³ rad/s
I = Moment of inertia = 0.00125 kg·m²
ω = Angular speed = 2010.619 rad/s
τ = 9.696×10⁻¹³ rad/s × 0.00125 kg·m² × 2010.619 rad/s =
2.44 ×10⁻¹² kg·m²/ s² = 2.44 ×10⁻¹² N·m
The required torque is 2.44 ×10⁻¹² N·m.
Answer:
a) 138 units
b) 17 units
c) 17 units
d) Total Cost = $353.35
Explanation:
Given:
Average pizzas delivered = 200
Charge of inventory holding = 30% of cost
Lead time = 7 days
Now,
a) Economic Order Quantity =
also,
Annual Demand = 200 × 12 = 2400
Cost per Order = Cost of Box + Processing Costs
= 30 cents + $10
= $10.30
and, Carrying Cost =
=
=
= $2.575
Therefore,
Economic Order Quantity =
= 138.56 ≈ 138 units
b) Reorder Point
= (average daily unit sales × the lead time in days) + safety stock
= (
= 46.67 ≈ 47 units
c) Number of orders per year =
=
= 17.39 ≈ 17 units
d) Total Annual Cost (Total Inventory Cost)
= Ordering Cost + Holding Cost
Now,
The ordering Cost = Cost per Order × Total Number of orders per year
= $10.30 × 17
= $175.1
and,
Holding Cost = Average Inventory Held × Carrying Cost per unit
Average Inventory Held = = 69
Carrying Cost per unit = $2.575
Holding Cost = 69 × $2.575 =
$177.675
Therefore,
Total Cost = Ordering Cost + carrying cost
= $175.1 + $177.675 = $353.35