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Triss [41]
3 years ago
12

rite stepwise equations for protonation or deprotonation of this polyprotic species in water. What are the products of the first

step? CO32- + H2O → __ + __ CO32- + H2O →
Chemistry
1 answer:
adoni [48]3 years ago
3 0

<u>Answer:</u> The products of the given chemical equation are HCO_3^-\text{ and }OH^-

<u>Explanation:</u>

Protonation equation is defined as the equation in which protons get added in the substance.

The chemical equation for the protonation of carbonate ion in the presence of water follows:

CO_3^{2-}+H_2O\rightarrow HCO_3^-+OH^-

By Stoichiometry of the reaction:

1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion

Hence, the products of the given chemical equation are HCO_3^-\text{ and }OH^-

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Given the following balanced equation, determine the rate of reaction with respect to [SO2].
Gnoma [55]

Answer:

Rate = -1/2 Δ[SO<sub>2</sub>]/Δt

so its gonna be (in more simple terms) rate= -1/2Δ(SO2)/Δt

Explanation:

7 0
2 years ago
The density of carbon in the form of diamond is 3.51 g/cm^3. if you have a small diamond with a volume of 0.0270 cm3 what is its
lora16 [44]
Hey there!:


density  = 3.51 g/cm³

Volume = 0.0270 cm³

Therefore:

D = m / V

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m = 3.51 * 0.0270

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3 0
3 years ago
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3 years ago
What type of energy is the sun of an objects potential and kinetic energy
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6 0
3 years ago
The equilibrium constant (K p) for the interconversion of PCl 5 and PCl 3 is 0.0121:
aksik [14]

Answer: At equilibrium, the partial pressure of PCl_{3} is 0.0330 atm.

Explanation:

The partial pressure of PCl_{3} is equal to the partial pressure of Cl_{2}. Hence, let us assume that x quantity of PCl_{5} is decomposed and gives x quantity of PCl_{3} and x quantity of Cl_{2}.

Therefore, at equilibrium the species along with their partial pressures are as follows.

                         PCl_{5}(g) \rightarrow PCl_{3}(g) + Cl_{2}(g)\\

At equilibrium:  0.123-x          x              x

Now, expression for K_{p} of this reaction is as follows.

K_{p} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}\\0.0121 = \frac{x \times x}{(0.123 - x)}\\x = 0.0330

Thus, we can conclude that at equilibrium, the partial pressure of PCl_{3} is 0.0330 atm.

4 0
3 years ago
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