To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.
The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L
Rearranging for T1:
T1 = (V1 × T2) / V2
Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C
The L- isomer serves as the enantiomer of the D- isomer, and given that the optical rotation of the D- isomer is + 13.5°, the L- isomer's optical rotation will have the same magnitude but an opposite sign, resulting in -13.5°.
Thus, the rotation of the racemic mixture will be equal to 0°.
- This occurs because a racemic mixture contains equal proportions of both enantiomers.
Answer:
A pressure of 137.14 MPa exists 10,000 m beneath the ocean surface.
At this same depth, the density measures 2039 kg/m3.
Explanation:
P0 and ρ0 symbolize the pressure and density at sea level (indicative of atmospheric conditions). With an increase in ocean depth, both pressure and density likewise rise.
The relationship between pressure and density can be expressed as:
By rearranging
This equation allows for computation of P at 10,000 m beneath the ocean's surface:
The density found at a depth of 10,000 m in the ocean is
To calculate the moles of MgSO4.7H2O, we find the molar mass equals 246, thus moles = 32 / 246 = 0.13 moles. Upon heating, all 7 H2O from one molecule will evaporate. The total moles of H2O present amount to 7 x 0.13 = 0.91, and the mass of that H2O is 0.91 x 18 = 16.38g. Therefore, the mass of the anhydrous MgSO4 that remains is 32 - 16.38 = 15.62 g.
Answer:
The temperature of the gas rises.
Explanation:
This is classified as an ISOCHORIC process where the volume remains unchanged. There is no work done by the system.
The gas only receives internal energy from the heat transferred to it from the surroundings.
In this situation, the pressure also increases.
