Answer: The stick model method
Equilibrium equation is
<span>Ag2CO3(s) <---> 2 Ag+(aq) + CO32-(aq) </span>
<span>From the reaction equation above, the formula for Ksp: </span>
<span>Ksp = [Ag+]^2 [CO32-] = 8.1 x 10^-12 </span>
<span>You know [CO32-], so you can solve for [Ag+] as: </span>
<span>(8.1 x 10^-12) = [Ag+]^2 (0.025) </span>
<span>[Ag+]^2 = 3.24 x 10^-10 </span>
<span>[Ag+] = 1.8 x 10^-5 M </span>
Answer:
1.85 × 10⁻⁶
Explanation:
0.0003 ÷ 162 = 1.851851852 × 10⁻⁶ ⇒ 1.85 × 10⁻⁶
Hope that helps.
Answer:
Amount of Ca(NO3)2 produced = 14.02 g
Explanation:
The given reaction can be depicted as follows:
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O
Since it is given that HNO3 is in excess, the limiting reactant is Ca(OH)2
Now, Mass of Ca(OH)2 = 6.33 g
Molar mass of Ca(OH)2 = 74 g/mol
Based on the reaction stoichiometry:
1 mole of Ca(OH)2 forms 1 mole of Ca(NO3)2
Therefore, moles of Ca(NO3)2 produced from the moles of Ca(OH)2 reacted = 0.0855 moles
Molar mass of Ca(NO3)2 = 164 g/mol
Answer:
Options A and B are correct.
Sodium salts and Carbon based fuels satisfy the criteria.
Explanation:
Sodium salts and Carbon Based fuels produce bright yellow emissions that can mask other emission colors.
Magnesium Salts produce no emissions, although, Magnesium metal burns brightly white while Aluminium salts produce white emissions.
Hope this Helps!!!
