It's "C" a sample of dust particles at 0 Pascals
Answer:
The new pressure is 53.3 kPa
Explanation:
This problem can be solved by this law. when the volume remains constant, pressure changes directly proportional as the Aboslute T° is modified.
T° increase → Pressure increase
T° decrease → Pressure decrease
In this case, temperature was really decreased. So the pressure must be lower.
P₁ / T₁ = P₂ / T₂
80 kPa / 300K = P₂/200K
(80 kPa / 300K) . 200 K = P₂ → 53.3 kPa
The solution would be like this for this specific problem:
Given:
pH of a 0.55 M hypobromous
acid (HBrO) at 25.0 °C = 4.48
[H+] = 10^-4.48 = 3.31 x
10^-5 M = [BrO-] <span>
Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>
To add, Hypobromous Acid does not require acid
adjustment, which is necessary for chlorine-based product and is stable and
effective in pH ranges of 5-9.<span>
</span>Hypobromous Acid combines with organic
compounds to form a bromamine. Chlorine also combines with the same organic
compounds to form a chloramine. <span>It is also
one of the least expensive intervention antimicrobial compounds available.</span>
