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mina [271]
3 years ago
6

Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super

script of X and Y, fill in the remaining blanks below by writing the appropriate subscript for each X and Y to balance the net charge on the ionic XaYb compound formed (where a and b represent positive, whole numbers of X and Y, respectively).
Y- Y2- Y3-
X+
X2+ XY2
X3+
Chemistry
1 answer:
dmitriy555 [2]3 years ago
8 0

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below:X^{m+}Y^{n-}\rightarrow X_nY_m

Thus, for all the given combinations, we obtain:

- Y⁻

X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3

- Y²⁻

X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3

- Y³⁻

X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY

Best regards!

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How many moles of solute are in 300 mL of 1.5 M CaCl2? How many grams fo CaCl2 is this?
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Answer:

1. 0.45 mole

2. 49.95g

Explanation:

The following were obtained from the question:

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1. We can obtain the mole of the solute as follow:

Molarity = mole of solute /Volume of solution

1.5 = mole of solute/0.3

Mole of solute = 1.5 x 0.3

Mole of solute = 0.45 mole

2. The grams in 0.45 mole of CaCl2 can be obtained as follow:

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Mole of CaCl2 = 0.45 mole

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What is the mass, in grams, of 0.0490 mol of iron(III) phosphate
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Answer:

m = 7.39 g.

Explanation:

Hello!

In this case, since the molar mass of iron (III) phosphate is 150.82 g/mol based on its molecular formula (FePO₄), we can compute the mass in grams of 0.0490 moles of this compound by setting up the following dimensional analysis:

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