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>
A. 1.01 is the right answer
Since
The formula is Pv= nRT
P=1 atm
V= 22.4 L
N= x
r= 0.0821
t = 273 k (bc it’s standard temperature)
So (1)(22.4)=(x)(0.0821)(273)
X= 1.001
Answer:
16mL
Explanation:
Using the following formula;
CaVa = CbVb
Where;
Where
Ca = concentration/molarity of acid (M)
Va = volume of acid (mL)
Cb = concentration/molarity of base (M)
Vb = volume of base (mL)
According to the information provided in this question;
Ca (HCl) = 2M
Cb (NaOH) = 5M
Va (HCl) = 40mL
Vb (NaOH) = ?
Using CaVa = CbVb
Vb = CaVa/Cb
Vb = 2 × 40/5
Vb = 80/5
Vb = 16mL
Answer:
C i took the test as well and i reambee i lern this in class
Explanation:
Hope it helps and mark me brainlest answer
