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Vadim26 [7]
3 years ago
10

Consider the chemical equation for the ionization of CH3NH2 in water. Estimate the percent ionization of CH3NH2 in a 0.050 M CH3

NH2(aq) solution. (Kb for CH3NH2 = 4.4 × 10-4)
Chemistry
1 answer:
anyanavicka [17]3 years ago
8 0

Answer:  The percent ionization of CH_3NH_2 in a 0.050 M CH_3NH_2(aq) solution is 8.9 %

Explanation:

CH_3NH_2+H_2O\rightarrow OH^-+CH_3NH_3^+

 cM                            0             0

c-c\alpha                       c\alpha            c\alpha

So dissociation constant will be:

K_b=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= concentration = 0.050 M and \alpha = degree of ionisation = ?

K_b=4.4\times 10^{-4}

Putting in the values we get:

4.4\times 10^{-4}=\frac{(0.050\times \alpha)^2}{(0.050-0.050\times \alpha)}

(\alpha)=0.089

percent ionisation =0.089\times 100=8.9\%

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laiz [17]

Answer:

B. 1.65 L

Explanation:

Step 1: Write the balanced equation

2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)

Step 2: Calculate the moles of SO₂

The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol

Step 3: Calculate the moles of SO₃ produced

0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃

Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP

At STP, 1 mole of an ideal gas occupies 22.4 L.

0.0736 mol × 22.4 L/1 mol = 1.65 L

6 0
3 years ago
A student constructs a galvanic cell that has a strip of iron metal immersed in a solution of 0.1M Fe(NO3)2 as one half-cell and
MrMuchimi

Answer:

Fe

Explanation:

The cell potential is:

ΔE°cell = E°red(red) - E°red(oxid)

Where, E°red(red) is the reduction potential of the substance that is reducing, and E°red(oxid) is the reduction potential of the substance that is oxidizing. For the reaction be spontaneous and happen, ΔE°cell > 0.

The reduction takes place in the cathode, which is the negative pole, and the oxidation in the anode, which is the positive pole. So, the electrons flow from the positive pole to the negative pole (anode to cathode).

Then, if the voltmeter measured a negative potential, it means that is was attached incorrectly. So, the anode is Fe.

4 0
3 years ago
Suppose a solution was too concentrated for an accurate reading with the spectrophotometer. The concentrated solution was dilute
yan [13]
Using the law of <span>dilution:

</span>initial Molarity = 3.5x10⁻⁶ M

<span>Initial volume = 4.00 mL
</span>
final Molarity = ??

final volume = 1.00 mL

Therefore:

Mi x Vi = Mf x Vf

(3.5x10⁻⁶) x 4.00 = Mf x 1.00

1.4x10⁻⁵ = Mf x 1.00

Mf = 1.4x10⁻⁵ / 1.00 = 

 1.4x10⁻⁵ M 


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True, but where is the question?
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