Answer:
Hey hi
Explanation:
Can you pls tell me which language is this pls I really want to help you... Sorry
The answer is strong winds
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>
<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
<h2>pH = 8.34</h2>
Answer:
81 °C
Explanation:
This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:
q = heat added/released by a sample
m = mass of sample
c=specific heat of sample
ΔT = change in temperature
from here we can rearrange the equation to state:
q/(mc) = ΔT
1200J/((20.0g)(4.2J/g•°C)) = ΔT
14°C = ΔT
If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.
B and e
first we need to balance the NH3 hence first we do E and multiplying the coefficient by 2. that will leave us with N2+H2–>2NH3.
N2 and H2 is balanced and now all that is left to do is the balance H2 by 3 as there is 6H on RHS hence we need 6H on LHS
