Answer:
6.76 moles.
Explanation:
2CO(g)+O2 (g) =2CO2(g)
When 2 CO mols were reacted with excess O2 then 2 mols of CO2 is created.
Therefore if 6.76 moles reacted, same number of CO2 will be created.
50 grams or 50,000 mili grams is the mass of solute in 1000 grams of a solution having a concentration of 5 parts per million.
Explanation:
Total mass of solution = 1000 grams or 1000 ml since 1 gram = 1 ml
concentration is 5 parts per million ( 5 mg in 1000 ml solution or 0.005 gram in 1000 ml)
the formula used for parts per million:
parts per million = 
putting the values in the equation:
parts per million = 
0.005 x 1000 = mass of solute
50 grams= mass of solute
converting this into mg
50,000 mg. is the total mass of solute in 5ppm of 1000 ml solution.
Answer:
Mass of sodium chloride decomposed = 24.54 g
Explanation:
Given data:
Mass of sodium chloride decomposed = ?
Mass of chlorine gas formed = 15 g
Solution:
Chemical equation:
2NaCl → 2Na + Cl₂
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 15 g/ 71 g/mol
Number of moles = 0.21 mol
Now we will compare the moles of Cl₂ with NaCl from balance chemical equation.
Cl₂ : NaCl
1 : 2
0.21 : 2×0.21 = 0.42 mol
Mass of Sodium chloride decompose:
Mass = number of moles × molar mass
Mass = 0.42 mol × 58.44 g/mol
Mass = 24.54 g
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.
