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Sladkaya [172]
3 years ago
12

What needs to happen for a solute to dissolve in a particular solvent

Chemistry
1 answer:
sergiy2304 [10]3 years ago
7 0
<span>the atractions between the solute and solvent molecules must be greater than the atractions keeping the solute together and the atractions keeping the solvent togetherrr.</span>
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How many iodide ions are present in 65.5ml of .210 m AlI3 solution
blondinia [14]

Answer:

2.48\times 10^{22} ions are present in solution.

Explanation:

Molarity of the solution = 0.210 M

Volume of the solution = 65.5 ml = 0.0655 L

Moles of aluminum iodide= n

Molarity=\frac{\text{Moles of compound}}{\text{Volume of the solution(L)}}

0.210M=\frac{n}{0.0655 L}

n = 0.013755 moles of aluminum iodide

1 mole of aluminum iodide contains 3 moles of iodide ions:

Then 0.013755 moles of aluminum iodide will contain:

3\times 0.013755 moles=0.041265 mol of iodide ions

Number of iodide ions in 0.041265 moles:

0.041265 mol\times 6.022\times 10^{23}=2.48\times 10^{22} ions

2.48\times 10^{22} ions are present in solution.

7 0
4 years ago
What mineral is used to create iron ore
Sav [38]
It would be gold ...................
8 0
4 years ago
Diazomethane has the molecular formula CH2N2. Draw the preferred Lewis structure for diazomethane and assign formal charges to a
Alja [10]

Answer:

See explanation

Explanation:

We define the formal charge on an atom in a molecule as the charge it carries assuming that electrons in all chemical bonds of the molecule were shared equally between atoms irrespective of the electronegativity of each atom.

The formula for calculating the formal charge on an atom in a molecule is;

Formal Charge = [number of valence electrons on neutral atom] – [(number of lone electron pairs) + (½ number of bonding electrons)] ·

The formal charge on the two nitrogen atoms in diazomethane is obtained as follows;

Middle nitrogen atom = 5 – 8/2 – 0 = +1

Last nitrogen atom = 5 – 4/2 – 4 = –1

The Lewis structure of the molecule is shown in the image attached.

8 0
3 years ago
Equal volumes of two solutions, one containing a strong acid at pH 2 and the other containing a strong base at pH 12, are mixed.
mash [69]

Answer:

7  

Explanation:

Assume we have 1 L of each solution.

Solution 1

\text{[H$^{+}$]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}

Solution 2

pH = 12

pOH = 14.00 - pOH = 14.00 - 12 = 2.0

\text{[OH$^{-}$]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}

3. pH after mixing

               H⁺  +  OH⁻ ⟶ H₂O

I/mol:     10⁻²    10⁻²  

C/mol:   -10⁻²   -10⁻²

E/mol:      0        0

The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.

pH = 7

8 0
4 years ago
A gas with a volume of 4.0L at a pressure of 205kPa is allowed to expand to a volume of 12.0L. What is the pressure in the conta
mr Goodwill [35]

Answer:

68.3 kilo Pascal is the pressure in the container.

Explanation:

To calculate the new pressure , we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=205kPa\\V_1=4.0 L\\P_2=?\\V_2=12.0 L

Putting values in above equation, we get:

P_2=\frac{P_1V_1}{V_2}

=\frac{205 kPa\times 4.0 L}{12.0 L}=68.3 kPa

68.3 kilo Pascal is the pressure in the container.

5 0
3 years ago
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