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4vir4ik [10]
3 years ago
11

what is the period of oscillation of an electron that is bouncing up and down in response to the passage of a packet of red lgih

t?
Chemistry
1 answer:
Talja [164]3 years ago
7 0

Answer:

2.33*10^-15 s

Explanation:

We need to estimate the period of oscillation of an electron that bounced up/down due to the effects of a passing packet of red light. The period of oscillation is the wavelength of the red light divided by the speed of the red light. We know that:

V = f*λ

where V = speed of the red light = 3*10^8 m/s

f = frequency = 1/T where T is the period of oscillation in seconds (s)

λ = the wavelength of the red light = 0.7*10^-6 m

Thus:

V = λ/T

T = λ/V = (0.7*10^-6)/(3*10^8) = 2.33*10^-15 s

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The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol
Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration PCl_3 and Cl_2.

\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}

\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M

\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}

\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M

The given equilibrium reaction is,

                            PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initially                 0.70        0.70              0

At equilibrium    (0.70-x)   (0.70-x)           x

The expression of K_c will be,

K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}

K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}

Now put all the given values in the above expression, we get:

49=\frac{(x)}{(0.70-x)\times (0.70-x)}

By solving the term x, we get

x=0.59\text{ and }0.83

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of PCl_3 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of Cl_2 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of PCl_5 at equilibrium = x = 0.59 M

Therefore, the concentration of PCl_3 at equilibrium is 0.11 M

3 0
3 years ago
What is the ph of a solution made by mixing 25.00 ml of 0.100 m hcl with 40.00 ml of 0.100 m koh? assume that the volumes of the
Digiron [165]
<span>Answer: The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first. moles HCl = 0.04000 L * 0.100 M = 0.00400 moles moles KOH = 0.02500 L * 0.100 M = 0.00250 moles moles HCl left = 0.00400 - 0.00250 = 0.00150 moles Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+] pH = -log [H+] = -log (0.0231) = 1.64</span>
3 0
3 years ago
At a certain temperature the vapor pressure of pure acetic acid HCH3CO2 is measured to be 226.torr. Suppose a solution is prepar
Rudiy27

Answer:

The partial pressure of acetic acid is 73.5 torr

Explanation:

Step 1: Data given

Total pressure is 226 torr

mass of acetic acid = 126 grams

mass of methanol = 141 grams

 

Step 2: Calculate moles of acetic acid

moles acetic acid = mass acetic acid / molar mass acetic acid

moles acetic acid = 127 grams / 60.05 g/mol

moles acetic acid = 2.115 moles

Step 3: Calculate moles of methanol

moles methanol = 141 grams / 32.04 g/mol

moles methanol = 4.40 moles

Step 4: Calculate total moles

Total moles = moles of acetic acid + moles methanol

Total moles = 2.115 moles + 4.40 moles

Total moles = 6.515 moles

Step 5: Calculate mole fraction of acetic acid

2.115 moles / 6.515 moles = 0.325

Step 6: Calculate partial pressure of acetic acid

P(acetic acid) = 0.325 * 226

P(acetic acid) = 73.45 torr ≈73.5

 

We can control this by calculating the partial pressure of methanol

mole fraction of methanol = (6.515-2.115)/6.515 = 0.675

P(methanol) = 0.675 * 226 = 152.55

226 - 152.55 = 73.45 torr  

The partial pressure of acetic acid is 73.5 torr

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