Answer:
V₂ = 111.3 mL
Explanation:
Given data:
Initial volume of gas = 50.0 mL
Initial temperature = standard = 273.15 K
Final volume = ?
Final temperature = 335 °C (335+273.15 = 608.15 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 50.0 mL ×608.15 K / 273.15 k
V₂ = 30407.5 mL.K / 273.15 K
V₂ = 111.3 mL
To determine the mass of CO₂, the following must be known :
- the molar mass of CaCO₃
- the mole ratio of CaCO₃ to CO₂
- the molar mass of CO₂
<h3>Further explanation</h3>
Reaction
Decomposition of CaCO₃
CaCO₃ ⇒ CaO + CO₂
Given the mass of CaCO₃, so to determine the mass of CO₂ :
1. Find the mol of CaCO₃ from the molar mass of CaCO₃

2. Find the mole ratio of CaCO₃ : CO₂(from equation = 1 : 1)

3. Find the mass of CO₂ from the molar mass of CO₂

Given information :
G = 173.3 KJ
H = 180.7 KJ
T = 303.0 K
S = unknown (?)
By using the given formula : G = H - TS , we can calculate the value of 'S'
On rearranging the formula we get : S = 
Plug in the value of G , H and T in the above formula :
S = 
S = 0.02442 
Answer:
4.76
Explanation:
In this case, we have to start with the <u>buffer system</u>:

We have an acid (
) and a base (
). Therefore we can write the <u>henderson-hasselbach reaction</u>:
![pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=pH~%3D~pKa%2BLog%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
If we want to calculate the pH, we have to <u>calculate the pKa</u>:

According to the problem, we have the <u>same concentration</u> for the acid and the base 0.1M. Therefore:
![[CH_3COO^-]=[CH_3COOH]](https://tex.z-dn.net/?f=%5BCH_3COO%5E-%5D%3D%5BCH_3COOH%5D)
If we divide:
![\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D~%3D~1)
If we do the Log of 1:

So:

With this in mind, the pH is 4.76.
I hope it helps!
Answer: 568g/mol
Explanation:
It should be noted that there are 40 atoms of carbon in lycopene.
Since mass of 1 carbon = 12g/mol
Mass of 40 carbon atoms = 40 × 12g/mol = 480g/mol
Let the molar mass of lycopene be represented by x.
Therefore the molar mass of carbon = x × mass percent of carbon in lycopene
x × 84.49% = 480g/mol
x × 0.8449 = 480g/mol
x = 480/0.8449
x = 568g/mol
The molar mass of lycopene is 568g/mol