Answer:
True
Explanation:
The physical changes are reversible in most cases and these changes are not the chemical changes which means that it is only the change in its state not in their nature. Just take the example of water, on cooling it becomes solid and change in color can be seen which is white in solid form and colorless in liquid form. This is also reversible and is a physical change. This means that physical changes can be identified at macroscopic level. Hence the answer is true.
The decomposition time : 7.69 min ≈ 7.7 min
<h3>Further explanation</h3>
Given
rate constant : 0.029/min
a concentration of 0.050 mol L to a concentration of 0.040 mol L
Required
the decomposition time
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time
For first-order reaction :
[A]=[Ao]e^(-kt)
or
ln[A]=-kt+ln(A0)
Input the value :
ln(0.040)=-(0.029)t+ln(0.050)
-3.219 = -0.029t -2.996
-0.223 =-0.029t
t=7.69 minutes
Carbon dioxide and oxygen are removed from the air.
Explanation:
When air is passed through aqueous sodium hydroxide solution the carbon dioxide is removed from the air.
First the carbon dioxide will dissolve and react with water to form carbonic acid ( H₂CO₃) :
CO₂ + H₂O → H₂CO₃
The the carbonic acid will react with sodium hydroxide to form sodium carbonate (Na₂CO₃):
H₂CO₃ + 2 NaOH → Na₂CO₃ + 2 H₂O
After this by passing the air over heated cooper the oxygen is removed.
2 Cu + O₂ → 2 CuO
Learn more about:
neutralization reaction
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Explanation:
The given data is as follows.
Concentration = 0.1 
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= 
T =
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
= 
where,
= concentration = 
Hence, putting the given values into the above equation as follows.
=
=
=
m
or, =
= 1 nm (approx)
Also, it is known that
= 
Hence, we can conclude that addition of 0.1
of KCl in 0.1
of NaBr "
" will decrease but not significantly.