Barium can be analyzed by precipitating it as bas04 and weighing the precipitate. when a 0.713-g sample of a barium compound was treated with excess h2s04, 0.5331 g of baso4 formed. what is the percentage of barium in the compound?
molecular mass of baso4 233.39 g/mol
molecular mass of ba = 137.327 g/mol

31.37%

Explanation:

For this case, you should consider the following reaction:

Ba⁺²₍aq₎ + H₂SO₄ ₍aq₎ → BaSO₄ ₍s₎ + H₂O

For which you obtain the precipitate of BaSO₄

In order to obtain the mas of Barium on the precipitate, you may use the following formula:

gBa= M₍BaSO₄₎x(M₍Ba₎/M₍BaSO₄₎)

Where:

gBa= mass of Barium

M₍BaSO₄ ₎= mass of BaSO₄ from the precipitate

M₍Ba₎= mass of Barium from the original sample

M₍BaSO₄₎= mass of BaSO₄ from the precipitate

gBa= (0.5331)x(137.327/233.39)= 0.3136 g

Then we ontain the percentage of Barium multiplying by 100:

% Ba on the original sample= 31.36%

the child will have heterozygous dominant chromosomes as characters of his father are dominant over the characters of his mother.

the child is heterozygous

state the principle of separating two immiscible liquids by separating funnel. describe an activity with diagram to separate a mixture of water and kerosene oil. answer: immiscible layers separate out in layers depending on their densities in separating funne