Answer:
The pressure changes from 2.13 atm to 1.80 atm.
Explanation:
Given data:
Initial pressure = ?
Final pressure = 1.80 atm
Initial temperature = 86.0°C (86.0 + 273 = 359 K)
Final temperature = 30.0°C (30+273 =303 K)
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
P₁ = P₂T₁ /T₂
P₁ = 1.80 atm × 359 K / 303 K
P₁ = 646.2 atm. K /303 K
P₁ = 2.13 atm
The pressure changes from 2.13 atm to 1.80 atm.
Answer:
(NH4)2S(aq) + Pb(NO3)2(aq) --> 2NH4NO3 (aq) + PbS (s)
<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.
<u>Explanation:</u>
All radioactive decay processes undergoes first order reaction.
To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:
![k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%20%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = ?
t = time taken = 1.52 hrs
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= Initial concentration of reactant = 100 g
[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g
Putting values in above equation, we get:
To calculate the half life period of first order reaction, we use the equation:
where,
= half life period of first order reaction = ?
k = rate constant = 
Putting values in above equation, we get:
Hence, the half life of the sample of silver-112 is 3.303 hours.
%yield = 88.5%
<h3>Further explanation</h3>
Given
Reaction
Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
Required
The percent yield
Solution
mol AgNO₃(MW=169,87 g/mol) :
= mass : MW
= 127 : 169.87
= 0.748
mol Ag from equation :
= 2/2 x mol AgNO₃
= 2/2 x 0.748
= 0.748
Mass Ag (theoretical) :
= mol x Ar Ag
= 0.748 x 108
= 80.784
% yield = (actual/theoretical) x 100%
%yield = 71.5/80.784 x 100%
<em>%yield = 88.5%</em>
