Answer:
D. Weathering includes the transportation of sediment while erosion does not.
The balanced chemical reaction would be:
<span>2NI3 = N2 + 3I2
We use the reaction above and the molar masses of the substances involved. We start with the initial amount of NI3 reactant.
3.58 g NI3 (1 mol NI3 / 394.71 g NI3) (3 mol I2 / 2 mol NI3) = 0.0136 mol I2
Therefore, option C is the answer.</span>
Answer:
The relative humidity is 36%.
Explanation:
Relative humidity is the measure of ratio of actual vapor density to saturated vapor density in a given system. So there are different ways to measure the relative humidity in the atmosphere. One of the common ways is to measure using Psychrometer. In this instrument, two different thermometers are used as measuring device. One of the thermometer measures the humidity in dry air and another in wet air. So the difference of the temperature of dry and wet air bulb will be related to determine the relative humidity percentage. The formula to determine the relative humidity using this method is as below.
![Relative humidity=\frac{(e_{w}-[N*(1+0.00115*T_{w}*(T_{d}-T_{w}))])*100}{e_{d} }](https://tex.z-dn.net/?f=Relative%20humidity%3D%5Cfrac%7B%28e_%7Bw%7D-%5BN%2A%281%2B0.00115%2AT_%7Bw%7D%2A%28T_%7Bd%7D-T_%7Bw%7D%29%29%5D%29%2A100%7D%7Be_%7Bd%7D%20%7D)
Also,
![e_{d}=6.112*e^{[\frac{(17.502*T_{d} )}{(240.97+T_{d} )}]}](https://tex.z-dn.net/?f=e_%7Bd%7D%3D6.112%2Ae%5E%7B%5B%5Cfrac%7B%2817.502%2AT_%7Bd%7D%20%29%7D%7B%28240.97%2BT_%7Bd%7D%20%29%7D%5D%7D)
And,
![e_{w}=6.112*e^{[\frac{(17.502*T_{w} )}{(240.97+T_{w} )}]}](https://tex.z-dn.net/?f=e_%7Bw%7D%3D6.112%2Ae%5E%7B%5B%5Cfrac%7B%2817.502%2AT_%7Bw%7D%20%29%7D%7B%28240.97%2BT_%7Bw%7D%20%29%7D%5D%7D)
Here e =2.718,
and
are the temperature at dry bulb and wet bulb temperatures, respectively.
So, the relative humidity after substituting the values of Td = 2 C and Tw = -2 C is 36%.
You have to google the answers
According to this formula :
ΔTf= i Kf m
i is van't Hoff factor= 1
Kf = 1.86
m the molality we need to assume it
m= x moles of C2H5OH / Kg of mass
∴ 15 = 1 * 1.86 * ( x moles of C2H5OH/ 0.45 kg)
∴X = 3.629 moles
mass = no.of moles x molar mass of C2H5OH
= 3.629 X 46 = 167 g
∴the volume = mass / denisty
= 167 / 0.7893 = 211.57≈ 212 mL