Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L
At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then
PV = c
Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.
It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,
P₁V₁ = P₂V₂
Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.
In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,
1.00 atm x 3.60 L = 2.50 atm x y
We cleared y to find V₂,
V₂ = y = 
= 1.44 L
Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>
Answer :
The concentration of 
before any titrant added to our starting material is 0.200 M.
The pH based on this ion concentration is 0.698
Explanation :
First we have to calculate the concentration of before any titrant is added to our starting material.
As we are given:
Concentration of HBr = 0.200 M
As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion and bromide ion 
.
As, 1 M of HBr dissociates to give 1 M of
So, 0.200 M of HBr dissociates to give 0.200 M of
Thus, the concentration of before any titrant added to our starting material is 0.200 M.
Now we have to calculate the pH based on this ion concentration.
pH : It is defined as the negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
Thus, the pH based on this ion concentration is 0.698
Answer:
81 °C
Explanation:
This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:
q = heat added/released by a sample
m = mass of sample
c=specific heat of sample
ΔT = change in temperature
from here we can rearrange the equation to state:
q/(mc) = ΔT
1200J/((20.0g)(4.2J/g•°C)) = ΔT
14°C = ΔT
If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.
Answer:
The diagram on the right. It has increments of 0.1 of a unit and therefore will provide a more precise measurement. The diagram on the right measures 88.4. In terms of the diagram on the left, a decimal answer cannot be determined because the increments are too large. Therefore the diagram on the left is less precise.
