The correct answer to this question is "5." the oxidation number of cl in ClO3 will be a positive 5 because oxygen is naturally a -2 charge. times that by three and then account for the negative charge of the CLO3- ion.
From what i can gather it looks like d
It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:
Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH)
With the values put in: 0.35 x 45 = 0.35 x V(NaOH)
= 45 ml.
There is 45 ml of V(NaOH)
Let me know if you need anything else. :)
- Dotz
Answer:
Heyo (Ish Mash Potato) XD
Explanation:
a freighter carrying a cargo of uranium hexafluoride sank in the english channel in 1984. the cargo of uranium hexafluoride weighed 2.25x10 to the eight power. kg and was contained in 30 drums, each having a volume of 1.62x10 to the sixth power L. what is the density in g/ml, of uranium hexafloride.
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