Answer:
basic unit of le nervous system is ur nerves brother nah im jp ur brain bcs if u didnt have a brain u wouldnt feel anything :D
Explanation:
Answer:
A 12 oz Coca Cola contains 39g of sugar or C6H12O6.
To calculate for the molarity of sugar in the soda, convert 39 grams of sugar to moles sugar:
39g/ 180.16 g/mol = 0.216 mol sugar
then, convert 12 oz to L:
12oz / (1oz/0.02957L) = 0.35484 L
therefore the concentration of sugar in the soda is:
M = mol sugar / L sol'n
= 0.216 mol sugar / 0.35484 L
= 0.609 M
Explanation:
Answer:
Explanation:
In a chemical formula, the oxidation state of transition metals can be determined by establishing the relationships between the electrons gained and that which is lost by an atom.
We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.
The oxidation state is usually expressed using the oxidation number and it is a formal charge assigned to an atom which is present in a molecule or ion.
To ascertain the oxidation state, we have to comply with some rules:
- The algebraic sum of all oxidation numbers of an atom in a neutral compound is zero.
- The algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion.
For example, let us find the oxidation state of Cr in Cr₂O₇²⁻
This would be: 2x + 7(-2) = -2
x = +6
We see that the oxidation number of Cr, a transition metal in the given ion is +6.
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg
Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol
The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺
The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5
Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol
Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg
I would say is because of the Boron from O?
