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2 years ago

6

Can you show me the answer and explain?

1 answer:

Hitman42 [59]2 years ago

6 0

Answer:

C

Explanation:

looking at a periodic table X is fluorine and Y is potassium

Fluorine is in group 7 and forms a 1- charge (which gains electrons) and potassium is in group 1 and forms a 1+ charge (which loses electrons)

Fluorine (X) has an electronic structure of 2,7 and needs to gain an electron from Potassium (Y) to have a full outer shell and potassium has an electronic structure of 2,8,8,1 so needs to lose an electron to have a full outer shell as well. This means that the electron that potassium (Y) has lost is given away to fluorine (X), so both elements become stable.

This is known as ionic bonding where metals (like potassium) lose electrons and non-metals (like fluorine) gain electrons to become more stable, forming ions

Any further clarification let me know

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2 years ago

It takes 412. KJ/mol to break a carbon-hydrogen single bond. Calculate the maximum wavelength of light for which a carbon-hydrog

olga_2 [115]

Answer:

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7 0

3 years ago

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

$\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

$Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)$

3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

4 0

2 years ago