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Kruka [31]
2 years ago
6

CH3CN major species present when dissolved in water

Chemistry
2 answers:
mario62 [17]2 years ago
5 0
The major species present when CH3CN is dissolved in water are CH3+ and CN-. It doesn't completely dissociate in water since it is not a strong acid or strong base. Only a strong acid or base will dissociate completely in water. For example HCl dissociates into H+ and Cl-, and NaOH dissociates into Na+ and OH-.
Margaret [11]2 years ago
3 0

Answer: CH₃CN and H₂O.

Explanation:

1) The spieces present in a solution may be either the molecules, in case of covalent compounds, or ions, in case of ionic compounds that dissociate (ionize).

2) Both, CH₃CN and H₂O are covalent (polar covalent) substances, so they do not ionize and the spieces in the solution are the molecules per se.

3) In solution, the molecules of H₂O will solvate the molecules of CH₃CN, meaning that H₂O molecules are able to separate the molecules of CH₃N from each other, and so every molecule of CH₃CN will end surrounded by many molecules of H₂O.

This happens because the interaction between the polar molecules of the two different compounds is strong enough to overcome the intermolecular forces between the molecules of the same compound.

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Naya [18.7K]

Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter

Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.

The algebraic expression would be:

<em>ppb [=] micrograms of compound/liter of solution</em>

We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.

For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.

The respective procedure is in a attached file.  

5 0
2 years ago
Describe the changes in properties (from metals to nonmetals or from nonmetals to metals) as we move (a) down a periodic group a
defon

Answer:

Explanation:

The changes in properties from metals to non-metals on a periodic table can be measured and determined by the metallicity or electropositivity of elements.

Metallicity is a measure of the tendency of atoms of an element to lose electrons.

a.

Down a periodic group, metallicity increases.

b.

Across a period from left to right electropositivity or metallicity decreases.

Metals are found in the left part of periodic table and the most reactive metal sits in the lower left corner. Non-metals are towards the right side of the table.

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2 years ago
In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par
Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

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O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

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2 years ago
In two or more complete sentences explain how to balance the chemical equation and classify its reaction type. ___P4 + ___O2 ⟶ _
Serhud [2]

According to the conversation of mass, mass cannot be created or destroyed. This means whatever is done to one side, must be done to the other.

There are 4 Phosphorus atoms on the left, there must be 4 on the right. To do this, you must multiply the P2O3 by 2 to get 4 Phosphorus atoms and 6 Oxygen atoms. Now to balance the Oxygen atoms, you must multiply the oxygen atoms on the left by 3.

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2 years ago
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Analyze and solve this partially completed galvanic cell puzzle. There are 4 electrodes each identified by a letter of the alpha
klasskru [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  E_{cell}__{AC}} = 0.94

Explanation:

  From the question we are told that

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From the data give we can see that

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i.e           1.56 - 1.53 = 0.03

   In the same way we can say that

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=>        E_{cell}__{AC}}=1.56- 0.62

                       E_{cell}__{AC}} = 0.94

       

             

5 0
2 years ago
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