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:
0.171 M
Explanation:
Step 1: Given data
- Mass of H₃PO₄ (solute): 3.35 g
- Volume of solution (V): 200 mL
Step 2: Calculate the moles of solute
The molar mass of H₃PO₄ is 97.99 g/mol.
3.35 g × 1 mol/97.99 g = 0.0342 mol
Step 3: Convert "V" to liters
We will use the conversion factor 1 L = 1000 mL.
200 mL × 1 L/1000 mL = 0.200 L
Step 4: Calculate the molarity of the solution
We will use the definition of molarity.
M = moles of solute / liters of solution
M = 0.0342 mol/0.200 L = 0.171 M
Energy released from cell respiration is know as ATP
Answer:
2.99 M
Explanation:
In order to solve this problem we need to keep in mind the definition of molarity:
- Molarity = moles of solute / liters of solution
In order to calculate the moles of solute, we <u>convert 125.6 g of NaF into moles</u> using its <em>molar mass</em>:
- 125.6 g NaF ÷ 42 g/mol = 2.99 mol NaF
As the volume is already given, we can proceed to <em>calculate the molarity</em>:
- Molarity = 2.99 mol / 1.00 L = 2.99 M
Answer:
this reaction is an oxidation reaction