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torisob [31]
3 years ago
7

What is the oxidation state of cr in CrPO4 and Cr3(PO4)2?

Chemistry
2 answers:
soldi70 [24.7K]3 years ago
6 0

Answer:

Explanation:

Oxidation state of Cr in CrPO₄

As a general rule, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero.

The compound above is in its neutral state and we sum all the oxidation numbers and equate to zero:

The oxidation number of P is -2

                                           O is -2

Let the oxidation number of Cr be x:

                              x + (-2) + 4(-2) = 0

                              x -2-8 = 0

                               x -10 = 0

                                x = +10

For Cr in Cr₃(PO₄)₂

         Using the same rule:

        2(x) + 2[-2 + 4(-2)] = 0

           2x + 2(-2-8) = 0

            2x -20 = 0

                x = +10

Alborosie3 years ago
6 0

Answer:

CrPO_4: Cr +3\\Cr_3(PO_4)_2: Cr+2

Explanation:

CrPO_4 and Cr_3(PO_4)_2 are compounds known as ternary salts

This means that they are formed by a metal or a non-metal and an anion

 Its formula is M_x(X_yO_z)_w, that is, the cation is written first and then the anion and the simplified charges are exchanged if possible.

The anion PO_4 has a load of -3: PO_{4}^{-3}

Let's look at the first compound CrPO_4 we observe that when exchanging the charges 3 of thePO_4 does not appear therefore the charges are simplified as the charges are completely simplified, it means that the chromium has the same valence (numerically but with opposite sign) that the anion

Therefore the oxidation state of Cr in CrPO_4 is +3

Let's look at the second compound Cr_3(PO_4)_2, it is observed that when exchanging the valences, the 3 of the PO_4 is with the chromium, and with the anion is  2

As valencia are not multiples, they cannot be simplified.

When exchanging the valences, thePO_4 has the valence corresponding to the chromium which in this case is + 2

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A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained. How many
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<u>Answer: </u>

A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed

<u>Explanation:</u>

Given, the initial value of the sample, A_0 = 150mg

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The amount left after first half life will be ½.

The number of half-life is calculated by the formula

\frac{A}{A_{0}}=\left(\frac{1}{2}\right)^{N}

where N is the no. of half life

Substituting the values,

\frac{18.75}{150}=\left(\frac{1}{2}\right)^{N}

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On equating, we get, N = 3

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7 0
3 years ago
Gas has a volume of 247.3 ML and is at 100 Celsius and 745 Hg. If the mass of the gas is 0.347 g what is the molar mass of the v
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Answer:

The molar mass of the vapor is 43.83 g/mol

Explanation:

Given volume of gas = V = 247.3 mL = 0.2473 L

Temperature = T = 100^{\circ}C = 373 K

Pressure of the gas = P = 745 mmHg  (1 atm = 760 mmHg)

P = \displaystyle \frac{745}{760} \textrm{ atm} = 0.9802 \textrm{ atm}

Mass of vapor = 0.347 g

Assuming molar mass of gas to be M g/mol

The ideal gas equation is shown below

\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\0.98026 \textrm{ atm}\times 0.2473 \textrm{ L} = \displaystyle \frac{3.47 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 373\textrm{K} \\M = 43.834 \textrm{ g/mol}

The molar mass of the vapor comes out to be 43.834 g/mol

4 0
3 years ago
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