1st one= CaO +H2O=Ca(OH)2
product side-
Ca=1
O=2
H=2
Reactant side-
Ca=1
O=2
H=2
The first one is balanced for you
There is 1 calcium on each side 2 oxygens on each side and 2 hydrogens on each side
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.


There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6
And a water molecule, this is called a dehydration synthesis. when 2 molecule combine, a water molecule leave.
Answer: It takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.
Explanation:
Integrated rate law for second order kinetics is given by:
k = rate constant = 
= initial concentration = 0.860 M
a= concentration left after time t = 0.260 M

Thus it takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.
Explanation:
option no 4 is correct answer
I hope is helpful