Answer:
Explanation:
Hello!
In this case, since the average rate of reaction is computed as a change given by:
![r=\frac{\Delta [NH_4NO_2 ]}{\Delta t}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B%5CDelta%20%5BNH_4NO_2%20%5D%7D%7B%5CDelta%20t%7D)
In such a way, given the concentrations at the specified times, we plug them in to obtain:
Whose negative sign means the concentration decreased due to the decomposition.
Best regards!
Answer:
9.80 g
Explanation:
The molecular mass of the atoms mentioned in the question is as follows -
S = 32 g / mol
F = 19 g / mol
The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol
The mass of 6 F = 6 * 19 = 114 g /mol .
The percentage of F in the compound =
mass of 6 F / total mass of the compound * 100
Hence ,
The percentage of F in the compound = 114 g /mol / 146 g / mol * 100
78.08 %
Hence , from the question ,
In 12.56 g of the compound ,
The grams of F = 0.7808 * 12.56 = 9.80 g
Answer:
11.3 g
Explanation:
7.895 + 3.4 = 11.295
When rounded to correct number of significant figures --> 11.3
There are 3 significant figures in 11.3
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows
time 0 40 80 120 160
moles 0.100 0.067 0.045 0.030 0.020
Q1)
for the first 40 s change of moles ;
= -d[A] / t
= - (0.067-0.100)/40s
= 8.25 x 10⁻⁴ mol/s
for the next 40 s
= -(0.045-0.067)/40
= 5.5 x 10⁻⁴ mol/s
the 40 s after that
= -(0.030-0.045)/40 s
= 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction
rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction
Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation.
Since this is a first order reaction,
b = 1
therefore the reaction is
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
= 8.25 x 10⁻³ s⁻¹
