Explanation:
a) HNO2(aq) = HNO3(aq) + H2O(l) +NO(g)
b) SoCl2 (l) + H2O (l) = So2(g) + 2HCl(aq)
c) CH4 (g) + 2O2(g) = Co2 (g) + 2H2O(g)
d) 3CuO(s) + 2NH3 (g) = 3Cu(s) + 3H2O (l) + N2(g)
Wax is susceptible to heat. Wax is responds to heat addition. The forces in the wax when heat is added are being broken off and are much lesser as its original state. Hope this answers the question. Have a nice day.<span />
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>
Answer:
The coefficients are 2 for H₂O and 1 for Ca(OH)₂.
Explanation:
Let's consider the following reaction.
Ca(OH)₂(aq) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂O(l)
According to the balanced equation, the molar ratio of H₂O to Ca(OH)₂ is 2:1. Using this conversion factor, we have the following proportion:
moles Ca(OH)₂. (2 mol H₂O ÷ 1 mol Ca(OH)₂) = moles H₂O