Answer:
621.2090000000001 grams
Explanation:
1 moles Calcium to grams = 40.078 grams
15.5*40.078 = 621.2090000000001 g
Answer:
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Answer:
pH = 9.48
Explanation:
We have first to realize that NH₃ is a weak base:
NH₃ + H₂O ⇔ NH₄⁺ + OH⁻ Kb = 1.8 x 10⁻⁵
and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.
Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:
pOH = pKb + log ( [ conjugate acid ] / [ weak base ]
mol NH₃ = 0.139 L x 0.39 M = 0.054 mol
mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol
Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)
pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52
pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48
The solution is basic which agrees with NH₃ being a weak base.
Answer:
The value is 
Explanation:
From the question we are told that

The initial volume of the fluorocarbon gas is 
The final volume of the fluorocarbon gas is
The initial temperature of the fluorocarbon gas is 
The final temperature of the fluorocarbon gas is 
The initial pressure is 
The final pressure is 
Generally the equation for adiabatically reversible expansion is mathematically represented as
![T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }](https://tex.z-dn.net/?f=T_2%20%3D%20%20T_1%20%20%2A%20%5B%20%5Cfrac%7BV_1%7D%7BV_2%7D%20%5D%5E%7B%5Cfrac%7BR%7D%7BC_v%7D%20%7D)
Here R is the ideal gas constant with the value

So
=> 
Generally adiabatic reversible expansion can also be mathematically expressed as

=>
=> 
=>
So

=> 
Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.
Explanation : Given,
=21 torr/min
The balanced chemical reaction is,

The rate of disappearance of
= ![-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
The rate of disappearance of
= ![-\frac{d[Cl_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D)
The rate of formation of
= ![\frac{1}{2}\frac{d[NOCl]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D)
As we know that,
=21 torr/min
So,
![-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2021torr%2Fmin%3D10.5torr%2Fmin)
And,
![\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D)
![\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%3D21torr%2Fmin)
Now we have to calculate the rate change.
Rate change = Reactant rate - Product rate
Rate change = (21 + 10.5) - 21 = 10.5 torr/min
Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.