A material that allows heat/electricity to transfer is called a conductor.
Answer:
The chlorine gas and potassium bromide solution react to form liquid bromine and potassium chloride solution.
Explanation:
Chemical equation:
Cl₂(g) + KBr (aq) → KCl (aq) + Br₂(l)
Balanced chemical equation:
Cl₂(g) + 2KBr (aq) → 2KCl (aq) + Br₂(l)
This equation showed that the chlorine gas and potassium bromide solution react to form liquid bromine and potassium chloride solution.
Chlorine is more reactive than bromine it displace the bromine from potassium and form potassium chloride solution.
The given equation is balanced and completely hold the law of conservation of mass.
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
Explanation:
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Answer:a) 11.34 g of ethane
can be formed
b)
is the limiting reagent
c) 3.44 g of the excess reagent remains after the reaction is complete
Explanation:
To calculate the moles :
1. ![\text{Moles of} H_2=\frac{4.21}{2}=2.10moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20H_2%3D%5Cfrac%7B4.21%7D%7B2%7D%3D2.10moles)
2. ![\text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20C_2H_4%3D%5Cfrac%7B10.6%7D%7B28%7D%3D0.378moles)
According to stoichiometry :
1 mole of
require 1 mole of ![H_2](https://tex.z-dn.net/?f=H_2)
Thus 0.378 moles of
will require=
of ![H_2](https://tex.z-dn.net/?f=H_2)
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
moles of
left = (2.10-0.378) = 1.72 moles
mass of
left=![moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g](https://tex.z-dn.net/?f=moles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D1.72moles%5Ctimes%202g%2Fmol%3D3.44g)
According to stoichiometry :
As 1 mole of
give = 1 mole of ![C_2H_6](https://tex.z-dn.net/?f=C_2H_6)
Thus 0.378 moles of
give =
of ![C_2H_6](https://tex.z-dn.net/?f=C_2H_6)
Mass of ![C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g](https://tex.z-dn.net/?f=C_2H_6%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.378moles%5Ctimes%2030g%2Fmol%3D11.34g)
Thus 11.34 g of ethane is formed.