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3 years ago

What mass of K2SO4 would you measure out to prepare 550 ml of a 0.76 M solution

2 answers:

5 0

The molecular weight of K2SO4 is 174.26 g/mole. The mass of K2SO4 required to make this solution is calculated in the following way.
550mL * (0.76mole/1000mL) * (174.26g/mole) = 72.84gram
<span>I hope this helps.</span>

yulyashka [42]3 years ago

5 0

Answer:

$m_{K_2SO_4}=72.84gK_2SO_4$

Explanation:

Hello,

In this case, by considering that the both the volume of the solution of potassium sulfate and its concentration are given, one solves for the required mass considering the following proportional factors related stoichiometrically:

$m_{K_2SO_4}=550mLsln*\frac{1Lsln}{1000mLsln}*0.76\frac{molK_2SO_4}{Lsln}*\frac{174.259gK_2SO_4}{1molK_2SO_4}\\ m_{K_2SO_4}=72.84gK_2SO_4$

Best regards.

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Approximately 1 mL of two clear, colorless solutions-0.1 M Ba(NO3)2 and 0.1 M Na2SO4- were combined. Upon mixing, a thick, milky

murzikaleks [220]

Answer: balanced chemical equation: $Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Net ionic equation : $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced chemical equation is:

$Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

$Ba^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2Na^+(aq)+2NO_3^-(aq)$

The ions which are present on both the sides of the equation are sodium and nitrate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

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