Answer:
58.94 mL
Explanation:
V1 = 48.3 mL V2 = v mL
T1 = 22 degree celsius OR 295 k T2 = 87 degree celsius OR 360 k
We will use the gas equation:
PV = nRT
Since the Pressure (p) , number of moles (n) and the universal gas constant(R) are all constants in this given scenario,
we can say that
V / T = k , (where k is a constant)
Since this is the first case,
V1 / T1 = k --------------------(1)
For case 2:
Since we have the same constants, the equation will be the same
V / T = k (where k is the same constant from before)
V2 / T2 = k (Since this is the second case) ------------------(2)
From (1) and (2):
V1 / T1 = V2 / T2
Now, replacing the variables with the given values
48.3 / 295 = v / 360
v = 48.3*360 / 295
v = 58.94 mL
Therefore, the final volume of the gas is 58.94 mL
Answer:
2Mg^+ +O2 right arrow 2MgO
Explanation:
Answer : The final concentration of the seawater is, 2.909 mole/L
Explanation :
Formula used for osmotic pressure :
where,
= osmotic pressure = 70.0 bar = 70 atm
R = solution constant = 0.0821 Latm/moleK
T= temperature of solution = 
C = final concentration of seawater = ?
Now put all the given values in the above formula, we get the concentration of seawater.
Therefore, the final concentration of the seawater is, 2.909 mole/L
