B i think ye ghhhjhhjjjkxhb
Answer:
13.1dm³
Explanation:
Given parameters:
Pressure of gas = 880mmHg
760mmHg = 1 atm
880mmHg will give 1.16atm
Temperature = 303k
Mass of gas = 26.9g
Molar mass of CO₂ = 12 + 2(32) = 44g/mol
Number of moles = mass/molar mass = 26.9/44 = 0.61mole
Unknown:
Volume of the sample = ?
Solution:
To solve this problem, we use the ideal gas equation:
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant = 0.082atmdm³mol⁻¹k⁻¹
T is the temperature
V =
=
= 13.1dm³
Answer:
density=6.74g/ml
:320g÷47.5ml
d=6.74g/ml
thank you
<em><u>I </u></em><em><u>hope</u></em><em><u> </u></em><em><u>this </u></em><em><u>is </u></em><em><u>helpful</u></em>
Answer:
B
Explanation:
The Niels Bohr's atomic model superior to all the earlier models is because it showed how the electron could orbit the nucleus without falling into it
Answer:
Of the following equilibria, only one will shift to the right in response to a decrease in volume.
On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.
Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
Decrease the volume
If the volume of the container is decreased , the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. So, the equilibrium will shift in the direction number of gaseous moles are less.
On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.
On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.