First, let's start off by finding the mass of this whole hydrate.
(Note: the unit of measurement for mass will be amu)
Let's find the molecular mass of each element.




Now, let's find the mass of each compound.


We have 6 molecules of H2O, so multiply 18.015 by 6 then add that with the weight of CoCl2.


Now divide 108.09 (mass of all the H2O in the hydrate) by 237.923 (total mass of hydrate).


Turn that into a percentage and you get 45.431%.
Hope this helps! :)
Answer:
10.28 mol
Explanation:
S + 2O = SO2
(atm x L) ÷ (0.0821 x K)
(3.45 x 45.6) ÷ (0.0821 x 373)
=5.13726
Then round it to significant figures
=5.14
5.14 mol SO2 x (2 mol O ÷ 1 mol SO2)
=10.28 mol O
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.
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