Answer:
a) The heat which we supply to water during boiling is used to overcome these forces of attraction between the particles so that they become totally free and change into a gas. This latent heat does not increase the kinetic energy of water particles and hence no rise in temperature takes place during the boiling of water.
b) Steam produces more severe burns than boiling water even though both are at 100oC because steam contains more heat, in the form of latent heat, than boiling water.
Explanation:
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The film which you called it is photographic film and it is strip like sheet and it's surface is clrear and transparent plastic which is coated on
two sided one side is with gelatin emulsion consists microscopic small light-sensor which is actually very sensitive in nature silver halides crystals. The size shape and other characters of the crystals gives the sensitivity,contrast of different colors combination ,and resolution of the films.
Answer:
It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
Explanation:
Mass of gold = m
Volume of gold = v
Surface area on which gold is plated = 
Thickness of the gold plating = h = 0.5 mm = 0.05 cm
1 mm = 0.1 cm
Density of the gold = 
Moles of gold = 
According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :
of electrons
Number of electrons = N =
Charge on single electron = 
Total charge required = Q
Amount of current passes = I = 8 Ampere
Duration of time = T
It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
A calorimeter contains reactants and a substance to absorb the heat absorbed. The initial temperature (before the reaction) of the heat absorbent is measured and then the final temperature (after the reaction) is also measured. The absorbent's specific heat capacity and mass are also known. Given all of this data, the equation:
Q = mcΔT
To find the heat released.
The question is incomplete, here is the complete question:
At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas
The reaction is second order for 
with a rate constant of 
at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M
a) 1.01 b) 5.19 c) 0.299 d) 0.0880 e) 3.34
<u>Answer:</u> The time taken is 5.19 seconds
<u>Explanation:</u>
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant =
t = time taken = ?
[A] = concentration of substance after time 't' = 0.150 M
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.260 M
Putting values in above equation, we get:
Hence, the time taken is 5.19 seconds
