From the calculation, the molar mass of the solution is 141 g/mol.
<h3>What is the molar mass?</h3>
We know that;
ΔT = K m i
K = the freezing constant
m = molality of the solution
i = the Van't Hoft factor
The molality of the solution is obtained from;
m = ΔT/K i
m = 3.89/5.12 * 1
m = 0.76 m
Now;
0.76 = 26.7 /MM/0.250
0.76 = 26.7 /0.250MM
0.76 * 0.250MM = 26.7
MM= 26.7/0.76 * 0.250
MM = 141 g/mol
Learn more about molar mass:brainly.com/question/12127540?
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Answer:
what kind of chemistry is it going to be?
Answer:
C) It will accelerate.
Explanation:
According to Newton’s second law of motion, when an object is acted on by an unbalanced force, it will accelerate.
An unbalanced force will change the speed or direction (or both) of an object. A change in speed and/or direction is acceleration.
A) is wrong. The object will stop moving only if there is a balanced force in the opposite direction.
B) is wrong. The object will decrease speed only if the unbalanced force has a component opposite to the direction of motion.
d) is wrong. The object will increase speed only if the unbalanced force has a component in the direction of motion.
These problems are a bit interesting. :)
First let's write the molecular formula for ammonium carbonate.
NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)
17.6 gNH4CO3
Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.
17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)
Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)
NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol
17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4
Now just take the molar mass we found to convert that amount into moles!
4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4
<u>Answer:</u> The equilibrium concentration of 
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of = 2.00 M
The given chemical equation follows:
<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of is 0.332 M
