A divalent metal ion dissolved in dilute hydrochloric acid forms a precipitate when H2S is bubbled through the solution. Which i
1 answer:
Answer:
D. Cd2+
Explanation:
The metal ions are classified into four groups which is based on their solubility.
Group II cations are the cations which are based on the solubility of the Sulphides.
Group II () cations are the cations which produce very insoluble sulfides.
So, among the given options, is the only cation which comes in Group II and thus, it will precipitate.
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<u>Answer:</u> The new concentration of lemonade is 3.90 M
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of lemonade solution = 2.66 M
Volume of solution = 473 mL
Putting values in equation 1, we get:
Now, calculating the new concentration of lemonade by using equation 1:
Moles of lemonade = 1.26 moles
Volume of solution = (473 - 150) mL = 323 mL
Putting values in equation 1, we get:
Hence, the new concentration of lemonade is 3.90 M
<span>2 KClO3(s) → 3 O2(g) + 2 KCl(s)
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g