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solniwko [45]
3 years ago
7

A sample of ph3 has mass 27.5g how many molecules are in this sample

Chemistry
1 answer:
egoroff_w [7]3 years ago
6 0
12.34 is the answer I think
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The principle that in any closed system
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Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of
Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

5 0
4 years ago
Gina made a poster for plastic recycling week and included this information on her poster:
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Answer:

Gina Should Put Rubber Tires Under The  Synthetic Category

Gina Should Put Starch Under The Natural Category

Explanation:

Edge 2020

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3 years ago
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Which will contain the greater number of moles of potassium ion: 30.0 ml of 0.15 m k2cro4 or 25.0 ml of 0.080 m k3po4?
Dahasolnce [82]
<span>30.0 ml of 0.15 m K2CrO4 solution will have more potassium ions. Let's see the relative number of potassium ions for each solution. Since all the measurements are the same, the real difference is the K2CrO4 will only have 2 potassium ions per molecule while the K3PO4 solution will have 3 potassium ions per molecule. K2CrO4 solution 30.0 * 0.15 * 2 = 9 K3PO4 solution 25.0 * 0.080 * 3 = 6 Since 9 is greater than 6, the K2CrO4 solution will have more potassium ions.</span>
3 0
3 years ago
What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?
Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})

Hence, the probability of finding the particle in the  one-dimensional box is:

P = \int_{x_{1}}^{x_{2}} \psi^{2} dx

P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx

P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]

P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})    

Solving for n=4:

P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})    

P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})    

P = \frac{1}{8}

I hope it helps you!

7 0
3 years ago
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