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densk [106]
3 years ago
7

Arjun conducted an experiment in the field related to the rate of percolation. He observed that it took 10 min for 200 mL of wat

er to percolate through the soil sample. Calculate the rate of percolation
Chemistry
1 answer:
Lapatulllka [165]3 years ago
4 0

Answer:

dwdsdsds

Explanation:

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An isotope has a half-life of 10 minutes. after 20 minutes, what percentage of the original nuclei remain?
Elena-2011 [213]
To solve this, we can use two equations.
t1/2 = ln 2 / λ = 0.693 / λ  
   
where, t1/2 is half-life and λ is the decay constant.

t1/2 = 10 min = 0.693 / λ

Hence, λ = 0.693 / 10 min              -         (1)

Nt = Nο e∧(-λt)    
                
Nt = amount of atoms at t =t time
Nο= initial amount of atoms
t = time taken

by rearranging the equation,
Nt/Nο = e∧(-λt)                  -  (2)

From (1) and (2),

Nt/Nο = e∧(-(0.693 / 10 min) x 20 min) 
Nt/Nο = 0.2500

Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%
                                                     
= (Nt/Nο ) x 100%
                                                      = 0.2500 x 100%
                                                      = 25.00%

Hence, Percentage of remaining nuclei is 25.00%
6 0
2 years ago
What weather types affect earthworms?
torisob [31]
Usually when it rains the worms surface!

HOPE THIS HELPS!
4 0
3 years ago
A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
_ an electron moved around the nucleus in circular paths​
kifflom [539]

Answer: although it is convenient to think of the electron moving around the nucleus along circular paths

Explanation:

7 0
3 years ago
The main reason that cs2 has a higher boiling point than co2 is that cs2
OlgaM077 [116]

Explanation:

Inspite of having similar intermolecular forces, CS2 has a higher boiling point than CO2, since it has a greater molar mass. The potential energy of molecules reduces until a certain level as they get closer to each other. Although the polarity of both CO2 and CS2 are cancelled because of their linear structure.

4 0
3 years ago
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