0 ml of a 0.100 m solution of nh3 is titrated with 0.250m hcl. after 10.0 ml of the hcl has been added, the resultant solution i
1 answer:
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Answer:
The method is accurate in the calculation of the
Explanation:
As a first step we have to calculate the <u>average concentration </u>of find it by the method.
Then we have to find the<u> standard deviation:</u>
For the confidence interval we have to use the formula:
μ=Average±
Where:
t=t student constant with 95 % of confidence and 5 data=2.78
μ= ±
upper limit: 0.84
lower limit: 0.75
If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.
The answer for the following problem is mentioned below.
- <u><em>Therefore the final moles of the gas is 14.2 × </em></u>
<u><em> moles.</em></u>
Explanation:
Given:
Initial volume () = 230 ml
Final volume () = 860 ml
Initial moles () = 3.8 ×
moles
To find:
Final moles ()
We know;
According to the ideal gas equation;
P × V = n × R × T
where;
P represents the pressure of the gas
V represents the volume of the gas
n represents the no of the moles of the gas
R represents the universal gas constant
T represents the temperature of the gas
So;
V ∝ n
=
where,
() represents the initial volume of the gas
() represents the final volume of the gas
() represents the initial moles of the gas
() represents the final moles of the gas
Substituting the above values;
=
= 14.2 ×
moles
<u><em>Therefore the final moles of the gas is 14.2 × </em></u><u><em> moles.</em></u>