An ore sample is known to contain copper sulfate pentahydrate, CuSO4•5H2O. If a 10.000 g sample of the ore loses 0.8445 g of water when strongly heated, what is the maximum quantity of CuSO4•5H2O that could be in the sample

2,341g of CuSO₄.5H₂O

Explanation:

It is possible to assume all loses water comes from CuSO₄.5H₂O. Thus:

0,8445g H₂O (1mol H₂O / 18,015g) = 0,04688 moles of water.

As 1 mole of contains 5 moles of water, moles of CuSO₄.5H₂O are:

0,04688 moles H₂O × (1 mole CuSO₄.5H₂O / 5 moles H₂O) = 9,38x10⁻³ moles of CuSO₄.5H₂O

The molecular mass of CuSO₄.5H₂O is 249,69g/mol. Thus, mass of 9,38x10⁻³ moles of CuSO₄.5H₂O is:

9,38x10⁻³ moles of CuSO₄.5H₂O × (249,69g / mol) =

2,341g of CuSO₄.5H₂O is the maximum quantity of CuSO₄.5H₂O that could be in the sample.

I hope it helps!

answer: the figure is attached herewith.

step-by-step explanation:   we are given to sketch a unit circle and label the terminal points corresponding to

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.

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