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3 years ago

3Co2 (aq) 6NO3(aq) 6Na (aq) 2PO43–(aq) → Co3(PO4)2(s) 6Na (aq) 6NO3(aq) Identify the net ionic equation for this reaction.

Chemistry

1 answer:

Burka [1]3 years ago

8 0

Answer : The net ionic equation will be,

$3Co^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

$3Co^{2+}(aq)+6NO_3^-(aq)+6Na^+(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)+6Na^+(aq)+6NO_3^-(aq)$

In this equation, $NO_3^-(aq)\text{ and }Na^+(aq)$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

Thus, the net ionic equation will be,

$3Co^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)$

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Alkenes can be hydrated via the addition of borane to yield alcohols with non-Markovnikov regiochemistry.

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Answer:

The given statement is true.

Explanation:

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Then through the donation of a pair of electrons from the hydrogen peroxide ion, the process of oxidation takes place resulting in the formation of trialkylborane. After this realignment of an R group with its pair of bonding, electrons take place with adjacent oxygen resulting in the withdrawal of a hydroxide ion. Eventually, the trialkylborate reacts with the aqueous NaOH to generate alcohol and sodium borate as the side product.

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3 years ago

CaO + HCI → CaCl2 + H20

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Explanation:

To balance this chemical equation-

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Balanced form:-

CaO + 2 HCl ==. CaCl2 + H2O

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3 years ago

High concentrations of ammonia (NH3), nitrite ion, and nitrate ion in water can kill fish. Lethal concentrations of these specie

kykrilka [37]

Explanation:

It is known that molality is the number of moles present in kg of solution.

Mathematically, Molality = $\frac{\text{no. of moles of solute}}{\text{mass of solvent in Kg}}$

The given data is as follows.

Molar mass of ammonia = 17 g/mol

Concentration = 1.002 mg/L = $\frac{0.001002 g/L}{17 g/mol}$

= $5.89 \times 10^{-4}$ mol/L

Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Therefore, molality will be calculated as follows.

Molality = $\frac{5.89 \times 10^{-4} mol/L}{1 kg/L}$

= $5.89 \times 10^{-4} mol/kg$

And,

Molar mass of nitrite = 46 g/mol

Concentration = 0.387 mg/L = $\frac{0.000412 g/L}{46 g/mol}$

= $8.956 \times 10^{-6}$ mol/L

And, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality = $\frac{8.956 \times 10^{-6} mol/L}{1 kg/L}$

= $8.956 \times 10^{-6}$ mol/kg

Now, Molar mass of nitarte = 62 g/mol

Concentration = 1352.2 mg/L

= $\frac{1.3522 g/L}{62 g/mol}$

= 0.02181 mol/L

Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality will be calculated as follows.

Molality = $\frac{0.02181 mol/L}{1 kg/L}$

= 0.02181 mol/kg

Therefore, molality of given species is $5.89 \times 10^{-4} mol/kg$ for ammonia, $8.956 \times 10^{-6}$ mol/kg for nitrite, and 0.02181 mol/kg for nitrate ion.

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3 years ago

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Explain what the horizontal and vertical lines of a Fischer Projection indicate about the structure of the molecule

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Explanation:

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3 years ago