Answer:
CaBr2 is a compound because it is composed of two or more elemnts
Ca - Calcium
Br - Bromine
Br2 is not a compound because it is not two or more elements it is just one element.
Br - Bromine
Explanation:
The conversion factor is added to the original given unit so that you can end up with the final unit. Basically, the conversion unit does not change the value because the factor is just equal to 1. You just manipulate the units by cancelling them.
Among the given choices, the fraction which is equal to 1 is 10⁻³ g/ 1 mg, because there are 1,000 mg per 1 g. In reverse, that would be 10⁻³ of a gram.
Answer:
The problem is solvable
Explanation:
The information that is provided, together with some equations are enough to solve the problem:
- The inlet states are totally defined.
- A heat balance under adiabatic assumption, allows to calculate the outlet temperature (both outlets stream are in equilibrium, so at the same temperature).
- The rule of phases implies that por the two-phase equilibrium, there is only one pressure for each temperature.
- The mass balance and equilibrium relationships allows to calculate the res of properties for the outlet streams.
The balanced equation for the reaction is as follows;
Zn + 2HCl --> ZnCl₂ + H₂
stoichiometry of Zn to HCl is 1:2
limiting reagent is the reactant that is fully consumed in the reaction.
the amount of product formed depends on amount of limiting reactant present.
The excess reactant is when it has been provided in excess and only a fraction of the provided amount is used up in the reaction.
Amount of Zn moles - 0.5 mol
Amount of HCl moles - 0.6 mol
If Zn is provided, number of HCl moles reacted - 0.5 x 2 = 1.0 mol of HCl is required, but only 0.6 mol of HCl is provided.
HCl is therefore the limiting reactant,
Number of Zn moles required - 0.6 mol /2 = 0.3 mol
But 0.5 mol are given, therefore 0.2 mol of Zn is in excess
Answer:
hello your question is incomplete attached below is the complete question
answer :
Fe2O3 = 159 g/mol
CO = 28.01 g/mol
Fe = 55.85 g/mol
CO2 = 44.01 g/mol
Explanation:
Fe = 55.85 g/mol
O = 16.0 g/mol
hence the molar mass of Fe2O3 = 2(55.85) + 3(16) = 159 g/mol
C = 12.01 g/mol
O = 16.0 g/mol
hence the molar mass of CO = 28.01 g/mol
Fe = 55.85 g/mol
C = 12.01 g/mol
O = 16.0 g/mol
hence the molar mass of CO2 = 12.01 + 2(16) = 44.01 g/mol