We know that
1) empirical formula of a compound is the representation of a molecules with the elements in simplest mole ratio.
2) the molar mass is the actual proportion of moles of elements present in a molecule
3) there is a simple ratio between the molar mass and empirical mass a molecule
Now let us solve each problem
1) empirical formula of 
will be 
2) empirical formula of 
will be
3) empirical formula of 
will be 
4) the molecular formula will be :
the molecular formula = 4 X empirical formula = 4 X CO = 
5) the mole ratio of Cl and Cr is 
empirical formula will be 
Answer:
Yes.
Explanation:
"These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products."
There is a beginning product, and a reactant is needed in order for something to happen.
For example, according to Newton, something cannot happen until an exterior force comes and stops/pushes it.
Sorry if this is incorrect, I am just 4th grade :(
When a gas is heated in a sealed container, the gas particles move faster and hit the sides of the container with more force leading to an increase in the pressure of the gas.
An increase in the temperature of a gas increases the collision rate of the molecules of the gas according to the kinetic theory.
With increased kinetic energy, the molecules of the gases collide more frequently with each other and the walls of the container.
The collision creates an increase in the magnitude of pressure inside the container.
More on the kinetic theory of gases can be found here: brainly.com/question/15354399?referrer=searchResults
Answer:
To ensure provide and reliable data
Explanation:
Calibration ensures precise repeatable performance while preventing pipetting errors. Just the way measurement standards are established to distinguish valid and invalid processes, pipette calibration standards are designed to ensure the best pipette accuracy.
Answer:
23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr
Explanation:
The balanced equation here is
6NaBr + 1AlO3 = 3Na2O + 2AlBr3
6 moles of NaBr are required to produce 2 moles of AlBr3
Mass of one mole of NaBr = 102.894 g/mol
Mass of one mole of AlBr3 = 266.69 g/mol
Mass of 6 moles of NaBr = 6*102.894 g/mol
Mass of two moles of AlBr3 = 2*266.69 g/mol
6*102.894 g NaBr produces 2*266.69 g of AlBr3
23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr
