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2 years ago

2. Barium compounds are chemically very similar to strontium compounds. Water-soluble

2 answers:

6 0

The ability of sodium sulfate to act as an antidote to barium chloride poisoning is because displaces barium in barium chloride to form sodium chloride and insoluble harmless barium sulfate.

<h3>Why is a solution of sodium sulfate used as an antidote?</h3>

The ability of sodium sulfate to act as an antidote to barium chloride poisoning is because of the reactivity of sodium.

Sodium is a more reactive metal than barium, so it displaces barium in barium chloride to form sodium chloride and insoluble harmless barium sulfate.

The equation of the reaction is given below:

BaCl2 + Na2SO4 ----> BaSO4 + NaCl

Therefore, the ability of sodium sulfate to act as an antidote to barium chloride poisoning is because displaces barium in barium chloride to form sodium chloride and insoluble harmless barium sulfate.

Learn more about about reactivity of metals at: brainly.com/question/24866635

katovenus [111]2 years ago

4 0

There are different kinds of poisonings. Sodium sulfate are known to be an effective antidote because they create or produce a kind of precipitate with barium that is said to be not easily absorbed from the gastrointestinal tract.

There have been a lot of Cases of barium poisoning that are known to be followed by high hypokalemia.

Soluble sulfates such as sodium sulfate and others such as potassium infusion are known to be a key antidote.

If given promptly and in the right dosage, soluble sulfates are known to be an effective antidote.

Learn more about barium from

brainly.com/question/13656770

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A mixture of BaCl2 and NaCl is analyzed by precipitating all the barium as BaSO4. After addition of an excess of Na2SO4 to a 3.6

Juli2301 [7.4K]

Answer:

$\% m=40.46\%$

Explanation:

Hello there!

in this case, according to the given information, it turns out firstly necessary for us to write up the chemical equation as shown below:

$BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl$

Thus, we calculate the mass of BaCl2 stoichiometrically related to the produced 1.658 g of precipitate in order to discard it from the sample:

$m_{BaCl_2}=1.658gBaSO_4*\frac{1molBaSO_4}{233.38 gBaSO_4} *\frac{1molBaCl_2}{1molBaSO_4}*\frac{208.23 gBaCl_2}{1molBaCl_2}\\\\m_{BaCl_2}=1.479gBaCl_2$

Thus, the mass percentage is calculated as shown below:

$\% m=\frac{1.479g}{3.656g}*100 \% \\\\\% m=40.46\%$

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