Answer:
By atomic number?
Explanation:
fingers crossed its right :/
Answer:
Iron is the element that is produced at the limit of the reaction.
Explanation:
In nuclear fusion 2 lighter nuclei are combined together into a single nucleus releasing a tremendous amount of energy in the process.
Up to fusion of iron the reaction of fusion is exothermic but when iron atom pops out the reaction becomes endothermic and it requires very high amount of energy to fuse iron atoms which is not available thus marking an end to the fusion reaction.
M/V=D
65.14/35.4≈1.84
The density of the sulfuric acid would be about 1.84g/mL
<span>Answer:
Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either.
Start with your 2 half reactions:
I- --> IO3-
Cr2O72- --> 2 Cr3+
Balance O by adding H2O:
I- + 3 H2O --> IO3-
Cr2O72- --> 2 Cr3+ + 7H2O
Balance H by adding H+:
I- + 3 H2O --> IO3- + 6 H+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O
Balance charge by adding e-:
I- + 3 H2O --> IO3- + 6 H+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O
Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give:
Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
Answer:
The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M
Explanation:
We use the formulas:
pH= - log(H30+) and Kwater=(H30+)x(OH-)
pH= - log(H30+) ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M
Kwater=(H30+)x(OH-)
(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7