Answer:
host
Explanation:
the virus exploits your cells
Answer:
548 g/mol
Explanation:
The freezing point depression of a solvent occurs when a nonvolatile solute is added to it. Because of the interactions between solute-solvent, it is more difficult to break the bonds, so the phase change will need more energy, and the freezing point will drop, which is called cryoscopy.
The drop in temperature can be calculated by:
ΔT = Kf*W*i
Where Kf is the cryoscopy constant of the solvent, W is the molality, and i is the van't Hoff factor, which indicates the fraction of the solute that dissolves.
The molality represents how much moles (n) of the solute is presented in each kg of the solvent (m2), thus
W = n/m2
The number of moles is the mass of the solute (m1) in g, divided by the molar mass (M1) of it:
W = m1/(M1*m2)
So, by the data:
0.2214 = 0.632/(M1*0.00521)
0.00115M1 = 0.632
M1 = 548 g/mol
Phase transitions is when a substance go through a change in its physical state, for example, solid to liquid, liquid to gas, etc. Energy in this case must be absorbed or released, and not necessarily it will be seen a change of temperature in this case. Mathematically, we calculate the energy released or absorbed by using the molar enthalpy of vaporization or molar enthalpy of fusion, as we can see in the following formulas:
ΔH = m*ΔHvap
ΔH = m*ΔHfusion
So phase transition is the energy required for a substance to change the physical state.
Answer:
final volume = 10.5 Liters N₂(g) at 21°C and 823Torr*
Explanation:
*Note=>No specified mass value of N₂(g) is defined in the problem. Therefore for a starting point, the gas sample is assumed to be 1.00 mole N₂(g) at STP conditions 22.4L
Determine volume of N₂(g) at 21°C(=294K) and 823 Torr (= 2.286 Atm).
Start with Volume of N₂(g) at 0°C and 1 Atm pressure => 22.4L and adjust to final volume of N₂(g) based upon 21°C(=294K) and 823 Torr (= 2.286 Atm).
V(final) = 22.4L(294K/273K)(360 Torr/823 Torr) = 22.4L(294/273)(360/823) = 10.55 Liters final volume.
Note: The volume of 1 mole (assumed) of any gas at STP (0°C/1 Atm) is 22.4 Liters. To convert to non-STP conditions, convert temperature and pressure factors (changes) that reflect what happens when the gas is expanded or decreased; but, these adjustments are taken independently for each variable of interest. The following notes explain.
For the increase in temperature from 0°C(=273K) to 21°C(=294K) one must apply a temperature ratio that will increase volume. That is, the change in volume due to the temperature change is 294K/273K. If a 273K/294K ratio were used the volume would have decreased. Not so for heating a sample of gas.
For the increase in pressure one should expect a decrease in volume. Therefore apply a pressure ration that will effectively decrease the volume of the gas. That is, to decrease a 22.4L sample at STP multiply the standard volume by a ratio of pressures that will decrease 22.4L to a smaller volume. That is, V(final by pressure effects) multiply by 360Torr/823Torr to decrease the STP VOLUME (22.4L) to the new non-standard volume. If 823Torr/360Torr were used, the final volume would not be smaller, but larger. Such is the physical effect of an increasing pressure change.
Answer:
O2(g) was Reduced
Explanation:
It is necessary at this point to restate the working definitions of oxidation and reduction.
Oxidation refers to increase in oxidation number. A chemical specie is oxidized in a chemical reaction if there is a positive increase in its oxidation number from left to right in the reaction.
A chemical specie is said to be reduced when there is a decrease in its oxidation number from left to right in a reaction. Hence reduction refers to a decrease in oxidation number.
Now let us focus on O2(g). Its oxidation number on the left hand side is zero. On the right hand side, its oxidation number decreases to -2. This shows a decrease in oxidation number. From our premises above, we can safely conclude that O2(g) was reduced in the reaction.
