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pochemuha
3 years ago
14

An unknown metal sulfate is found to be 72.07% SO4 2- by mass. Assuming that the charge on the metal cation is +3, determine the

identity of the cation.
Chemistry
1 answer:
wariber [46]3 years ago
8 0
Lets name the unknown metal as M. Cation would be M³⁺.
the molecular formula of the compound is M₂(SO₄)₃
the mass of one mole - (molar mass of M x2 + 3 x molar mass of SO₄²⁻)
                                   = 2M + 96 x 3
                                   = 2M + 288
In 1 mol if there's 72.07% of sulphate , 
then 72.07 % corresponds to 288 g
               1 % is then      - 288/72.07 
       100 %  of the compound - 288/72.07 x 100 
 molar mass of the compound - 399.6 g/mol
mass of 2M - 399.6 - 288 = 111.6 g
molar mass of M - 111.6 /2 = 55.8 g/mol
the element with molar mass of 55.8 is Fe.
Unknown metal is iron(III) , Fe³⁺

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A pairs with T and G pairs with C.

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2. Which of these combinations of elements are most likely to react to form ionic compounds?
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b? prob

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Calculate the molar solubility of ca(io3)2 in each solution below. the ksp of calcium iodate is7.1 × 10−7.
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In order to find the answer, use an ICE chart:

Ca(IO3)2...Ca2+......IO3- 
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<span>less.......+x......+2x </span>
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</span>
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20.
sergiy2304 [10]
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A solution is prepared by mixing 525 mL of ethanol with 597 mL of water. The molarity of ethanol in the resulting solution is 8.
Alenkinab [10]

Answer:

\Delta V = 234.736\,mL

Explanation:

The quantity of moles of ethanol in the solution is:

n_{C_{2}H_{5}OH} = \left(\frac{597\,mL}{1000\,mL} \right)\cdot \left(8.35\,\frac{mol}{L} \right)

n_{C_{2}H_{5}OH} = 4.985\,mol

The mass and volume of ethanol in the solution are, respectively:

m_{C_{2}H_{5}OH} = (4.985\,mol)\cdot \left(46.07\,\frac{g}{mol} \right)

m_{C_{2}H_{5}OH} = 229.658\,g

V_{C_{2}H_{5}OH} = \frac{229.658\,g}{0.7893\,\frac{g}{mL} }

V_{C_{2}H_{5}OH} = 290.964\,mL

The difference between the total volume of water and ethanol mixed to prepare the solution and the actual volume of solution is:

\Delta V = (525\,mL+597\,mL) - (597\,mL + 290.964\,mL)

\Delta V = 234.736\,mL

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