1. HCl (H:1, Cl 35.5) ---> 1+35.5 = 36.5
2. FeS (Fe:56, S:32) ---> 56+32 = 88
3. Cl2 (Cl:35.5) ---> 35.5 x 2 = 71
4. CaC03 (Ca:40, C: 12, O: 16) ---> 40 + 12 + 3(16) = 100
5. Fe0 (Fe: 56, O: 16) ---> 56+16=72
Answer:
the molarity of NaOH is 1.10. And the molarity of HCl is 1.10. And the initial Temp=0.50(°c). and The final Temp= 1.10(°c)
Answer:
pH = 4.34
Explanation:
pH= -1/2(logKa) -1/2(log C)
= -1/2( log 5.98*10^-8) -1/2(log 0.0353)
=-1/2(-7.22)-1/2(-1.45)
=3.61+0.725= 4.34
A covalent bond? Not sure how much detail you want, sorry