Answer:
1.12g/mol
Explanation:
The freezing point depression of a solvent for the addition of a solute follows the equation:
ΔT = Kf*m*i
<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>
<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>
<em>m is molality of the solution</em>
<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>
<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>
Replacing:
26.22°C = 5.35°Ckgmol⁻¹*m*1
4.90mol/kg = molality of the compound X
As the mass of the solvent is 100g = 0.100kg:
4.9mol/kg * 0.100kg = 0.490moles
There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:
0.551g / 0.490mol
= 1.12g/mol
<em>This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.</em>
MgH2 + 2 H2O → Mg(OH)2 + 2 H2
Answer:
The final pressure is 0.788 atm (option b).
Explanation:
Boyle's law says that the volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure. That is: if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases. This is expressed mathematically as the product of pressure times volume equal to a constant value:
P*V=k
Assuming a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment, by varying the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:
P1*V1=P2*V2
In this case:
- P1= 2.14 atm
- V1= 3 L
- P2= ?
- V2= 8.15 L
Replacing:
2.14 atm*3 L= P2* 8.15 L
Solving:
0.788 atm= P2
<u><em>The final pressure is 0.788 atm (option b).</em></u>
Answer: The volume of the balloon at the center of the typhoon is 41.7L.
Note: The complete question is given below;
If a small weather balloon with a volume of 40.0 L at a pressure of 1.00 atmosphere was deployed at the edge of Typhoon Odessa, what was the volume of the balloon when it reached the center?
The severity of a tropical storm is related to the depressed atmospheric pressure at its center. In August 1985, Typhoon Odessa in the Pacific Ocean featured maximum winds of about 90 mi/hr and pressure that was 40.0 mbar lower at the center than normal atmospheric pressure. In contrast, the central pressure of Hurricane Andrew (pictured) was 90.0 mbar lower than its surroundings when it hit south Florida with winds as high as 165 mi/hr.
Explanation:
Since no temperature changes were given, it is assumed to be constant. Therefore, Boyle's law which describes the relationship between pressure and volume is used to determine the new volume at the center of Typhoon Odessa. Mathematically, Boyle's law states that; P1V1 = P2V2
Assuming 1atm = 1 bar, 1mbar = 0.001atm, 40mbar = 0.040atm
P1 = 1.0atm, V1 = 40.0L, P2 = 1atm - 0.040atm = 0.960atm, V2 = ?
Using P1V1 = P2V2
V2 = P1V1/P2
V2 = 1.0 * 40.0 / 0.96
V2 = 41.67L
Therefore, the volume of the balloon at the center of the typhoon is 41.7L.
The reaction is
2H₂(g) + O₂(g) ---> 2H₂O
Thus as per balanced equation two moles of hydrogen will react with one moles of oxygen.
There is a directly relation between moles and volume. [One mole of each gas occupies 22.4 L of volume at STP]
Thus we can say that two unit volume of hydrogen will react with one unit volume of oxygen
Now as we have started with equal units of volume of both oxygen and hydrogen, half of oxygen will be consumed against complete volume of hydrogen
so the gas which will remain in excess is oxygen
