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3 years ago

Explain why water and sodium oxide will have very different properties

Chemistry

1 answer:

Softa [21]3 years ago

4 0

Water and sodium oxide have different properties because of their nature as explained below.

<h3><u>Explanation:</u></h3>

Sodium oxide is a oxide of metallic sodium, while water is an oxide of hydrogen. So sodium oxide is a metallic oxide, while water is a non metallic oxide. Sodium oxide is a basic oxide, while water is neutral. As state of matter is concerned, sodium oxide is solid in normal room temperature, while water is liquid in normal room temperature. Water is a polar covalent molecule with partial charges on oxygen, but sodium oxide is an ionic molecule.

So all these factors contribute to very different properties of both sodium oxide and water.

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Is 6.022x 10^23 carbon atoms a mol

Kamila [148]

Answer:

Yes

Explanation:

1 mol of any substance is equal to the avagadro constant

8 0

3 years ago

An ionic compound has a highly negative ΔH soln in water. Would you expect it to be very soluble or nearly insoluble in water? E

valina [46]

The ionic compound will be soluble in water.

Ions are present in ionic compounds, which are held together by the attraction interactions between the ions with opposing charges. One of the most well-known ionic compounds is table salt or sodium chloride. Molecular compounds are made up of separate molecules that are connected by sharing electrons (covalent bonding).

High melting points are found in ionic substances. ionic compounds are brittle and rigid. When an ionic chemical is dissolved in water, it separates into ions. Ionic compound solutions and melting forms of these substances carry electricity, while solid materials do not.

A substance's heat of solution value is one of the criteria that determine whether or not it will dissolve. The Ionic compound will dissolve in water because a negative heat of the solution indicates that the reaction will be spontaneous.

To know more about Ionic compound refer to: brainly.com/question/9167977

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8 0

2 years ago

A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L

Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the <u>sodium azide (NaN₃)</u> present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)

The pH of this solution is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60$

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

$\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles$

Now, the number of moles of HN₃ is:

$\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles$

Then, the pH of the buffer solution after the addition of HCl is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43$

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

6 0

3 years ago

What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as

serious [3.7K]

CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

<span>Cl- = 0.4 moles (ANSWER)</span>

6 0

3 years ago

What is the systematic name of mg(no3)2

Oksi-84 [34.3K]

The systematic name of Mg(NO3)2 is magnesium nitrate. If you were confused, there are two of the nitrate because magnesium has a charge of 2+ while just one nitrate has a 1- charge. They would have to cancel out because it is an ionic compound.

7 0

4 years ago

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