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Lelechka [254]
3 years ago
10

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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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

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

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

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

7 0
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

5 0
3 years ago
Read 2 more answers
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