Hey, lovely! It's a pretty lengthy process but here is a pretty clear video on how to do it. Hope this helps ya!
https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/balancing-chemical-equat...
Answer:
frequency = 0.47×10⁴ Hz
Explanation:
Given data:
Wavelength of wave = 6.4× 10⁴ m
Frequency of wave = ?
Solution:
Formula:
Speed of wave = wavelength × frequency
Speed of wave = 3 × 10⁸ m/s
Now we will put the values in formula.
3 × 10⁸ m/s = 6.4× 10⁴ m × frequency
frequency = 3 × 10⁸ m/s / 6.4× 10⁴ m
frequency = 0.47×10⁴ /s
s⁻¹ = Hz
frequency = 0.47×10⁴ Hz
Thus the wave with wavelength of 6.4× 10⁴ m have 0.47×10⁴ Hz frequency.
There are 3 significant figures. Significant numbers are the numbers that build up your total number. 1-9 always count, 0 only counts if it’s after another number. For example: 0,901 has 3 significant numbers as does 0,910. 9,10 also has 3. 0,09 has just 1.
If you are provided with Cation and an Anion with different oxidation states, then there ratio in the formula unit is adjusted as such that the oxidation number of one ion is set the coefficient of other ion and vice versa,
Example:
Let suppose you are provided with A⁺² and B⁻¹, so multiply A by 1 and B by 2 as follow,
A(B)₂
In statement we are given with Co⁺³ and SO₄⁻², so multiply Co⁺³ by 2 and SO₄⁻² by 3, hence,
Co₂(SO₄)₃
Result:
Co₂(SO₄)₃ is the correct answer.
