Speed=distance/time
D=150km
S=7200s
Divide the numbers.
150/7200=0.02km/s <---final answer
We will use the expression for freezing point depression ∆Tf
∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is
∆Tf = 0C - (-3°C) = 3°C
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and one Cl- ion:
KCl → K+ + Cl-
Therefore, the molality m of the solution can be calculated as
3 = 2 * 1.86 * m
m = 3 / (2 * 1.86)
m = 0.80 molal
Answer: it is in the correct order
Explanation:
Answer:
Option A. the hydroxyl group (-OH)
Explanation:
Ethanol, CH₃CH₂OH belongs to the class of organic compound called alkanol.
They have general formula as R–OH
Where
R => is an alkyl group
OH => is the hydroxyl group
The hydroxyl group (OH) is the functional group of the alkanol (alcohol)
Answer:
At equilibrium, the concentration of 
is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
+ 
⇄ 2
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no nor present. Immediately, and are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: []=0 ; [ ]= 0 ; []=0.60M
C: []=+x ; [ ]= +x ; []=-2x
E: []=0+x ; [ ]= 0+x ; []=0.60-2x
Now we can use the constant information:
![K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5Bproducts%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D%7B%5Breactants%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D)
= 
= 
= 
At equilibrium, the concentration of is going to be 0.30M
