Answer:
It is present in the second group of the periodic table
Explanation:
Question:
a. a direct linear relationship
b. an inverse linear relationship
c. a direct nonlinear relationship
d. an inverse nonlinear relationship
Answer:
The correct option is;
d. An inverse nonlinear relationship
Explanation:
From the universal gas equation, we have;
P·V = n·R·T
Where we have the temperature, T and the number of moles, n constant, therefore, we have
P×V = Constant, because, R, the universal gas constant is also constant, hence;
P×V = C

Since P varies with V then the graphical relationship will be an inverse nonlinear as we have
V P C
1 5 5
2 2.5 5
3 1.67 5
4 1.25 5
5 1 5
6 0.83 5
7 0.7 5
8 0.63 5
9 0.56 5
10 0.5 5
Where:
V = Volume
P = Pressure
C = Constant = 5
P = C/V
The graph is attached.
Answer:
BK. BK. BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;

= 0.8
The rate-out
= 
= 
We can say that:

where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12

Integration of the above linear equation =

so we have:



∴ 
Since A(0) = 12
Then;



Hence;



∴ the concentration at 10 minutes is ;
=
%
= 0.0456667 %
= 0.046% to three decimal places
Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol