Assuming that the reactants are:
(NH4)2SO4 (aq) + Ba(NO3)2 (aq)
and the products are:
BaSO4 (s) + 2NH4NO3 (aq),
then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.
According to the solubility rules, the following elements are considered insoluble when paired with SO4:
Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+
Therefore, the precipitate will be BaSO4 (s).
They have no charge and are located inside the nucleus
1. by making a atomic configuration
2.by making a table of shells of k.l.m.n....
that's all
Answer:
V₂ = 520.42 mL
Explanation:
Given data:
Initial volume = 350.0 mL
Initial pressure = 840 mmHg
Initial temperature = 33°C (33 +273 = 306 K)
Final temperature = 52°C (52+273 = 325 K)
Final volume = ?
Final pressure = 600 mmHg
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 840 mmHg × 350.0 mL × 325 K / 306 K × 600 mmHg
V₂ = 95550000 mmHg.mL.K /183600 K.mmHg
V₂ = 520.42 mL
