Answer:
a minimum of <em>1</em><em>0</em><em>,</em><em>0</em><em>0</em><em>0</em><em> </em>years
Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
Answer:
The new concentration will be 0.01 M.
Explanation:
To determine the new concentration we use the following formula.
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.1 M
volume (1) = 100 mL
concentration (2) = unknown
volume (2) = 100 mL + 900 mL = 1000 mL
concentration (2) = [concentration (1) × volume (1)] / volume (2)
concentration (2) = (0.1 × 100) / 1000 = 0.01 M
