: A chemical process of decomposition involving the splitting of a bond by the addition of water.
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
False because that doesn’t make sense
Answer:
M
Explanation:
Concentration of ![Cu^{2+} = [Cu(NO_3)_2]](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%20%3D%20%5BCu%28NO_3%29_2%5D)
= 0.020 M
Constructing an ICE table;we have:
![Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%2B4NH_3_%7Baq%7D%20%5Crightleftharpoons%20%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D_%7B%28aq%29%7D)
Initial (M) 0.020 0.40 0
Change (M) - x - 4 x x
Equilibrium (M) 0.020 -x 0.40 - 4 x x
Given that: 
![K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}](https://tex.z-dn.net/?f=K_f%20%7D%20%3D%20%5Cfrac%7B%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D%7D%7B%5BCu%5E%7B2%2B%7D%5D%5BNH_3%5D%5E4%7D)
Since x is so small; 0.40 -4x = 0.40
Then:
M
Answer:
5.03 moles
Explanation:
Find the molar mass of C5H12 and you will get 72.17 g/mol
Next to find the number of moles, you divide 362.8 by the molar mass and you get
(362.8 g)/(72.17 g/mol)= 5.03 moles
