6 atoms of Cobalt.
4 atoms to Phosphate (PO4).
6 atoms to sodium.
6 atoms to Chlorine.
The coefficient is one factor that tells you how many atoms go to each atom/element, however, the subscript also influences this.
If the compound has parenthesis, then the subscript within the parenthesis remains untouched and does not affect the atoms.
Basically, to calculate the atoms is multiply the coefficient (number in front of the atom) by the subscript attached to the atom. If it's a compound such as Co3PO4, then make sure the coefficient affects the second atom as well.
Hope this helps!
Answer: B, the gas is being squeezed out of the liquid.
Explanation: The gas does not want to be trapped inside of the liquid, so it is trying to force it’s way out. Therefore creating more pressure in the container or whatever the liquid is being held in.
I would say #3 I’m sorry if it’s wrong tho
Answer:
70mol
Explanation:
The equation of the reaction is given as:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Given parameters:
Number of moles of acetylene = 35.0mol
Number of moles of oxygen in the tank = 84.0mol
Unknown:
Number of moles of CO₂ produced = 35.0mol
Solution:
From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.
To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:
From the equation:
2 moles of acetylene produced 4 moles of CO₂
∴ 35.0 mol of acetylene would produced:
= 70mol
Answer:
the energy of the third excited rotational state 
Explanation:
Given that :
hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm
Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.
Thus; the reduced mass μ = 
μ = 
μ = 
∵ 1 μ = 1.66 × 10⁻²⁷ kg
μ = 
μ = 1.6139 × 10⁻²⁷ kg

The rotational level Energy can be expressed by the equation:

where ;
J = 3 ( i.e third excited state) &




We know that :
1 J = 


