The acceleration is equal to force divided by mass which would be 100 / 50 which is 2 m/s2
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I think. Protons and electrons share the same number as well as the atomic number.
The integrated rate law expression for a first order reaction is
![ln\frac{[A_{0}]}{[A_{t}]}=kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D%3Dkt)
where
[A0]=100
[At]=6.25
[6.25% of 100 = 6.25]
k = 9.60X10⁻³s⁻¹
Putting values

taking log of 100/6.25
100/6.25 = 16
ln(16) = 2.7726
Time = 2.7726 / 0.0096 = 288.81 seconds
The osmotic pressure is a colligative property which depends upon the number of molecules and not the type of molecules
The relation between osmotic pressure and concentration is
πV = nRT
where
π = Osmotic pressure [ unit atm] = ?
V = volume
n = moles
R = gas constant = 0.0821 L atm / mol K
T = temperature = 20°C = 20 + 273.15 K = 293.15 K
also
Molarity = moles / Volume
So
Molarity = n/V = 5.5 M
Putting values
π = MRT
π = 5.5 X 0.0821 X 293.15 = 132.37 atm
Osmotic pressure of given glucose solution will be 132.37 atm
Answer:
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?
A) BrCH2CH2Br
B) CH3CH2CH2Br
C) CH3CHBr2
D) CH3CH2CH2CH3
E) BrCH2CH2CH2CH2Br
Explanation:
The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.
This is an example of free radical substitution.
The structure of ethane and its bromination is shown below:
Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).
Remaining all other products are possisble to form on free radical substitution of ethane.