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3 years ago

Enter the formula for the compound cobalt(II) phosphate. Express your answer as a chemical formula.

Chemistry

1 answer:

alexira [117]3 years ago

5 0

Answer:

$Co_3(PO_4)_2$

Explanation:

Naming of the ionic compounds:-

The name of the cation is written first and the the name of the anion is written after the name of the cation separated by single space.
The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
In case of transition metals, the oxidation state are written in roman numerals in bracket in front of positive ions.

Hence, given ionic compound:-

Cobalt(II) phosphate

So, Cobalt will have a positive charge of +2

Phosphate is $PO_4^{3-}$

So, the formula is :-

Co $PO_4^{3-}$

2 3

$Co_3(PO_4)_2$

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2 years ago

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$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

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Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

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3 years ago

Convert 13.4 degrees celcius into kelvin

Alenkasestr [34]

Answer:

286.55K

Explanation:

To convert to kelvin , add 237 .15

$13.4\°C + 273.15 = 286.55K$

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3 years ago

Read 2 more answers