Volume of Cl₂(g) is produced at 1.0 atm and 540.°C=4.5×10^4 L
As per the evenly distributed response
2NaCl (g) ----> 2Na(l)+ Cl2(g)
Calculate the amount of Cl2 that was formed as indicated below:
Moles of Cl2 = 31.0 kg of Na x (1000* 1 * 1 / 1*23* 2)
= 673.9 mol
P is equal to 1.0 atm, and T is equal to 813.15 K
when converted to Kelvin by multiplying by a factor of 273.15.
Using Cl2 as an ideal gas, determine the in the following volume:
volume = nRT/P
= 673.9 * 0.0821 * 813.15/ 1
=4.5×10^4 L
As a result, the volume of Cl2 under the given circumstances =4.5×10^4 L
Learn more about Volume here:
brainly.com/question/13338592
#SPJ4
You should classify it as co2 solid as CO2(s) and that small left comes to the bottom right hand side and dry ice is a form of liquid carbon dioxide so u should write it as CO2(l) and that L comes to the bottom right hand sise
Answer:
gold is metal which is a very good conductor for electric current and most chips are made out of it.
Actinium is the most reactive element or metal in group 3.
Explanation:
Reactivity of an element or metal is defined as its tendency to loose or gain atoms. Group 3 elements consist of metal. We know that reactivity of metal increases as we go down to the bottom of the column.
group 3 metals have 3 electrons in valence shell. They react by loosing 3 electrons. it has oxidation state of +3 and are electropositive.
Group 3 consists of scandium, yttrium, lanthanum and actinium.
Actinium is called actinide.
