Answer:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
Explanation:
We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.
Aqueous AgNO₃ and KI will dissociate in solution as follow:
AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)
KI (aq) —> K⁺(aq) + I¯(aq)
Aqueous AgNO₃ and KI will react as follow:
AgNO₃ (aq) + KI (aq) —>
Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)
Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.
Thus, the net ionic equation is:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
Answer:
all these are physical properties except release of heat so it's probably heat energy given off
You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
Answer:
3.816 × 10⁻³ M
Explanation:
<em>A stock solution of Cu²⁺(aq) was prepared by placing 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask and diluting to the mark with water. What is the concentration (in M) of Cu²⁺(aq)in the stock solution?</em>
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We can establish the following relations:
- The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
- 1 mole of Cu(NO₃)₂∙2.5H₂O contains 1 mole of Cu²⁺.
The moles of Cu²⁺ in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

The molarity of Cu²⁺ is:
