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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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What is a formula unit in an iconic bond

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Answer:

A formula unit indicates the lowest whole number ratio of ions in an ionic compound. Because the positively and negatively charged ions in an ionic compound are arranged in a crystal lattice, there is no discrete particle comparable to a molecule.

Explanation:

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If normal rainwater has a pH of about 6.50, what is the hydrogen ion concentration, [H+]?

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Answer:

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3 years ago

consider the balanced chemical equation below. when the chemical reaction was carried out calculated theoretical was yield for s

8_murik_8 [283]

Answer:

Explanation:

% yield = (actual yield / theoretical yield) X 100

For this question,

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7 0

3 years ago

purification of chromium can be achieved by electrorefining chromium from an impure chromium anode onto a pure chromium cathode

stiv31 [10]

The chromium in the electrolytic solution is present as cr³⁺ will take 6.19 hours will it take to plate 18.0 kg of chromium onto the cathode if the current passed through the cell is held constant at 45.0 A.

It is given that mass is 18 kg. Converting mass into grams

18 x (1000g / 1 kg)

= 18000 g

So, 18 kg = 18000 g

Now, calculate the number of moles

No. of moles = mass of Cr / molecular mass of Cr

No. of moles = 18000 / 52 g/mole

No. of moles = 346.15 mole

For , Cr³⁺, 3 moles of electrons are required

Hence,

3 x 346.15 mole

1038.45

As 96500 C of charge are present in 1 mole of electrons. As a result, the charge held by 1038.45 mole of electrons will be determined as follows.

= 1038.45mol x 96500 C / mole

= 10.02 x 10⁷ C

As, we know that relation between charge, current and time is as follows.

Q = I x T

Current is given as 45 A

Therefore, charge is calculated

T = Q / I

T = 100.21 x 10⁶ / 45

T = 2.227 x 10⁶

As there are 3600 seconds in one 1 hour. Converting 2.227 x 10⁶ into hours are

= 2.227 x 10⁶ / 3600 sec/h

= 6.19 hours

Thus, if the current flowing through the cell is maintained constant at 45 A, we may estimate that it will take 6.19 hours to plate 18 kg of chromium onto the cathode.

To know more about electrolytic solution visit :

brainly.com/question/9203208

#SPJ4

4 0

1 year ago

Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these were decomposed the sulfur dioxide produce

Zinaida [17]

Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

Explanation : Given,

Mass of oxygen in sulfur dioxide = 3.49 g

Mass of sulfur in sulfur dioxide = 3.50 g

Mass of oxygen in sulfur trioxide = 9.00 g

Mass of sulfur in sulfur trioxide = 6.00 g

Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.

Mass of oxygen per gram of sulfur for sulfur dioxide = $\frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}$

Mass of oxygen per gram of sulfur for sulfur dioxide = $\frac{3.49}{3.50}=0.997g$

and,

Mass of oxygen per gram of sulfur for sulfur trioxide = $\frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}$

Mass of oxygen per gram of sulfur for sulfur trioxide = $\frac{9.00}{6.00}=1.5g$

Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

8 0

3 years ago