The dimensions of room are . Thus, volume of room will be:
Converting to
Since,
Thus,
Now, concentration of carbon monoxide in the urban apartment is , thus, mass of carbon monoxide can be calculated as follows:
Rearranging,
Putting the values,
converting
Since, thus,
Therefore, mass of carbon monoxide present in the room is .
It is a graph. It shows observations and then you record your results with any of the graph types.
Answer:
121 K
Explanation:
Step 1: Given data
- Initial volume (V₁): 79.5 mL
- Initial temperature (T₁): -1.4°C
- Final volume (V₂): 35.3 mL
Step 2: Convert "-1.4°C" to Kelvin
We will use the following expression.
K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K
Step 3: Calculate the final temperature of the gas (T₂)
Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K
Answer:
(a) 5.04; (b) 5.18; (c) 12.30
Explanation:
(a) pH of buffer
HA + H₂O ⇌ H₃O⁺ + A⁻
(b) pH after addition of base
HA + H₂O ⇌ H₃O⁺ + A⁻
I/mol: 0.1 0 0.2
C/mol: -0.02 +x +0.02
E/mol: 0.08 x 0.22
(c) pH of water after addition of base
pOH} = -log0.02 = 1.70
pH = 14.00 - pOH = 14.00 - 1.70 = 12.30
1. You can use the formula V1/ T1 = V2/T2
Solve for V2 give than V1= 10.0 L, T1= 25 C= 298 K, T2= 10 C= 283 K.
(10.0L/298K)= V2/283K------> V2= 9.50 L.
I cant see the answer there.. weird!
2. you can use the formula P1V1= P2V2
solve for V2 given than P1= atm, V1= 2.0 L, P2= 300 atm.
(1atm)x(2.0L)= (300atm)V2----->0.00667 atm.
so answer is A.