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Dvinal [7]
3 years ago
13

Models may change as new discoveries are made. Please select the best answer from the choices provided T F

Chemistry
1 answer:
Serjik [45]3 years ago
8 0
True.
For an example, look at the models of the atom. First, we have the plumb pudding model. Then that was replaced by the planetary model. That model was replaced by the quantum model. 
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What is the total number of ions in one formula unit of (NH4)2CO3?
Len [333]

There are 3  ions in one formula unit of (NH4)2CO3.

Ions present in  Ammonium carbonate.(NH4)2CO3

(NH4)2CO3 is called Ammonium carbonate. The formula indicates that in one mole of ammonium carbonate, There are

  • Two moles of ammonium ions,NH₄⁺ with valency of +1.
  • 1 mole of Carbonate ions  CO₃²⁻  with valency of -2.

When 1 mole of ammonium carbonate is dissolved in water it gives

(NH₄)₂CO₃   ₊  H₂O---->2NH₄⁺    ₊  CO₃²⁻

Ions present are 2NH₄⁺   and CO₃²⁻

Learn more on (NH4)2CO3 here:brainly.com/question/2272720

4 0
2 years ago
Read 2 more answers
The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen: NO(g) + NO(g) → N2O2(g) (fast)
Karo-lina-s [1.5K]

Answer:

b. Second order in NO and first order in O₂.

Explanation:

A. The mechanism

\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)

B. The rate expressions

-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]

The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.

C. Assume the first step is an equilibrium

If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.

\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}

D. Substitute this concentration into the rate law

\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]

The reaction is second order in NO and first order in O₂.

8 0
3 years ago
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