Answer:
Pressure will decrease according to the Charles Gay Lussac law.
Explanation:
When there is a gas in a tank, and the moles of gas are not modified, neither the volume, we can notice that pressure will be modifying, in order to the absolute temperature, as directly proportion.
T° increase → Pressure increase
T° decrease → Pressure decrease
If the pressure keeps on constant, it will be the volume that would be modified according to the absolute T°, as a directly proportion.
Volume increase → T° increase
Volume decrease → T° decrease
Let's go to the maths
P₁ / T₁ = P₂/T₂
100 kPa / 333K = P₂ / 283K
(100 kPa / 333K) . 283K = P₂ → 84.9 atm
As T° has decreased, pressure also decreased.
Answer:- cell potential = -0.19 volts
Solution:- The equation that shows the connection between and cell potential, E is written as:
in this equation, n stands for moles of electrons, E stands for cell potential and F stands for faraday constant and it's value is .
It asks to calculate the value of E, so let's rearrange the equation:
Let's plug in the values in it:
since,
Where C stands for coulombs and V stands for volts.
So,
E = -0.19 V
So, the cell potential is -0.19 volts.
Answer:
It's B
Explanation:
I got it right on edjenuity
Answer:
V = 27.98 L
Explanation:
Given data:
Mass of CO₂ = 33.0 g
Pressure = 500 torr
Temperature = 27°C
Volume occupied = ?
Solution:
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 33.0 g/ 44 g/mol
Number of moles = 0.75 mol
Volume of CO₂:
PV = nRT
R = general gas constant = 0.0821 atm.L/ mol.K
Now we will convert the temperature.
27+273 = 300 K
Pressure = 500 /760 = 0.66 atm
By putting values,
0.66 atm×V = 0.75 mol × 0.0821 atm.L/ mol.K × 300 K
V = 18.47 atm.L/0.66 atm
V = 27.98 L
Answer:
The 3rd answer down.
Na²O (sodium oxide) will be a base when exposed to water H²O
Explanation:
Sodium Oxide Na²O, will become Sodium Hydroxide after being exposed to water (at 80% I believe).
The oxygen ion in Na²O has 2 extra electrons which makes it highly charged and very attractive to hydrogen ions. The attraction is so strong that when Na²O comes in contact with H²O, the O(-2) strips off a hydrogen from water, forming 2 x OH ions which of course are still strongly basic.