In order of increasing percent water content:CoCl₂.6H₂O, Ba(OH)₂.8H₂O, MgSO₄.7H₂O
<h3>Further explanation</h3>
CoCl₂.6H₂O.MW=237.90 g/mol
6H₂O MW = 6.18=108 g/mol
MgSO₄.7H₂O.MW=246.48 g/mol
MW 7H₂O = 7.18=126 g/mol
Ba(OH)₂.8H₂O MW=315.48 g/mol
MW 8H₂O = 8.18=144 g/mol
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.
Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.
Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.
Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:
Therefore, density of metal is 19.33 g/mL.
