Answer:
Tungsten is used for this experiment
Explanation:
This is a Thermal - equilibrium situation. we can use the equation.
Loss of Heat of the Metal = Gain of Heat by the Water
Q = mΔT
Q = heat
m = mass
ΔT = T₂ - T₁
T₂ = final temperature
T₁ = Initial temperature
Cp = Specific heat capacity
<u>Metal</u>
m = 83.8 g
T₂ = 50⁰C
T₁ = 600⁰C
Cp = 
<u>Water</u>
m = 75 g
T₂ = 50⁰C
T₁ = 30⁰C
Cp = 4.184 j.g⁻¹.⁰c⁻¹
⇒ - 83.8 x x (50 - 600) = 75 x 4.184 x (50 - 30)
⇒ = 
j.g⁻¹.⁰c⁻¹
We know specific heat capacity of Tungsten = 0.134 j.g⁻¹.⁰c⁻¹
So metal Tungsten used in this experiment
Answer:
Normality N = 0.2 N
Explanation:
Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.
Normality is represented by N.
Mathematically, we have :
Given that:
number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent
volume of solution (HCl) = 450 mL 450 × 10⁻³ L
Normality N = 0.2 N
The answer is
e o i and a
not really sure forgive me if im wrong
Explanation:
The given data is as follows.
T = 
= (120 + 273.15)K = 393.15 K,
As it is given that it is an equimolar mixture of n-pentane and isopentane.
So, 
= 0.5 and 
= 0.5
According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.
(393.15 K) = 9.2 bar
(393.15 K) = 10.5 bar
Hence, we will calculate the partial pressure of each component as follows.
= 
= 4.6 bar
and, 
= 
= 5.25 bar
Therefore, the bubble pressure will be as follows.
P = 
= 4.6 bar + 5.25 bar
= 9.85 bar
Now, we will calculate the vapor composition as follows.
= 
= 0.467
and, 
= 
= 0.527
Calculate the dew point as follows.
= 0.5, 
= 0.5
= 0.101966 
P = 9.807
Composition of the liquid phase is 
and its formula is as follows.
= 
= 0.5329
= 
= 0.467
Solve the following equation and check for extraneous solutions. Show all work. Thank you.
sqrt(sqrt(x - 3)) = sqrt(x - 15)
