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VashaNatasha [74]
3 years ago
6

What is the precipitate that will form from the reaction of barium chloride dihydrate with sodium sulfate decahydrate?

Chemistry
2 answers:
iragen [17]3 years ago
4 0
 <span>immediate milky appearance 
after 5-10 minutes a precipitate of BaSO4 will settle out of solution the the solution will be clear.
</span>but mostly a white precipitate

MakcuM [25]3 years ago
4 0

The formation of an insoluble salt on mixing of two solutions which contains soluble salts is refer to a precipitation reaction. The insoluble salt formed in the reaction is said to be a precipitate and can be separated by filtration or any other separation method.

The formula for barium chloride dihydrate is BaCl_2.2H_2O.

The formula for sodium sulfate decahydrate is Na_2SO_4.10H_2O.

The hydrates on the above compounds represents that they are in the aqueous phase. So, the reaction is:

BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + NaCl(aq)

The above equation is not balanced as the number of atoms of elements on the reactant side is not equal to the number of atoms of elements on the product side. So, in order to balance the reaction, NaCl on the reactant side is multiplied by 2. Thus, the balanced chemical equation is:

BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)

Hence, the precipitate that will form from the reaction of barium chloride dihydrate with sodium sulfate decahydrate is barium sulfate, BaSO_4.

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
azamat

Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

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(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

Best regards.

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So, on substituting we get,
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E_{eV} = 1.99 eV so, it can be rounded off to 2.00 eV.
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Answer:

H:H

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