The answer will be a or d im not really sure but the two of them will work
The partial pressure of methane in the mixture of methane and ethane has been 1 atm.
Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.
The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.
The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.
However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.
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Answer:
6.0 L
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 0.075 M
V1 = 200 L
M2 = 2.5 M
V2 = ?
Solve for V2 --> V2 = M1V1/M2
V2 = (0.075 M)(200 L) / (2.5 M) = 6.0 L
Answer:
Average speed of helium is higher than argon.
Explanation:
The average speed of helium is higher than argon atom under the conditions specified because of lower mass of the helium atom as compared to argon atom. Average speed of an atom is inversely proportional to mass of the atom. If mass of an atom decreases, the atom moves with higher speed while on the other hand, if the mass of an atom increases the average speed of an atom decreases.
First you have a knowledge of bond order which is
B.O=(no. of electrons in bonding orbital - no. of electrons in non-bonding orbital)÷2
Note:
bond strength is directly proportional to bond order.
For oxygen:
B.O=(6-2)/2= 2; after the removal of two electrons(removal occur from non-bonding orbital)
B.O=(6-0)/2= 3 (As B.O increased bond strength increased)
For Nitrogen:
B.O=(6-0)/2= 3; after the removal of two electrons(removal occur from bonding orbital)
B.O=(4-0)/2= 2 (As B.O decreased bond strength decreased)
