Answer:
The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Explanation:
Given the data in the question;
Using the Clapeyron equation
where 
is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu
T is the temperature ( 15 + 460 )R
m is the mass of water ( 0.5 Ibm )
is specific volume ( 1.5 ft³ )
we substitute
/ 
272.98 Ibf-ft²/R
Now,
where P₁ is the initial pressure ( 50 psia )
P₂ is the final pressure ( 60 psia )
T₁ is the initial temperature ( 15 + 460 )R
T₂ is the final temperature = ?
we substitute;
480.275 R
Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
The term saturated solution is used in chemistry to define a solution in which no more solute can be dissolved in the solvent. It is understood that saturation of the solution has been achieved when any additional substance that is added results in a solid precipitate or is let off as a gas.
Since the angle is West of North, therefore to find for
the westward component (horizontal component) of the vector, we use the sin
function:
sin θ = opposite side / hypotenuse = westward component /
resultant vector
So the westward component (x) is:
x = 85.42 sin 23
<span>x = 33.38 unit</span>
Answer:
Using the data in the table scientists, students, and others that are familiar with the periodic table can extract information concerning individual elements. For instance, a scientist can use carbon's atomic mass to determine how many carbon atoms there are in a 1 kilogram block of carbon.
Explanation:
HOPE THIS HELPS LIKE AN RATE PLZ
Answer:
= 8
Explanation:
noise from power mower = 106 dB
noise from rock concert = 115 dB
we would be applying the formula for the decibel level of sound to get the intensities.
(B) = 10 log \frac{I}{I₀}
I = intensity of sound
I₀= reference intensity = 10^{-12} W/m^{2}
- intensity of the power mower
106 = 10 log \frac{I}{10^{-12}}
10.6 = log \frac{I}{10^{-12}}
10^{10.6} = \frac{I}{10^{-12}}
I = 10^{10.6} x 10^{-12}
I of power mower = 0.0398 W/m^{2}
- intensity of the rock music
115 = 10 log \frac{I}{10^{-12}}
11.5 = log \frac{I}{10^{-12}}
10^{11.5} = \frac{I}{10^{-12}}
I = 10^{11.5} x 10^{-12}
I of rock music = 0.316 W/m^{2}
now the ratio of the intensity of the rock music to the power mower
= \frac{0.316}{0.0398}
= 8
