Answer:
by filtering it with filter paper
Answer:
6.022 ×10(index 23) / 7.5 = 0.8293 ×10(index 23)
Explanation:
molar mass of C = 12gmol
therefore in 12g of C there is one mole or an amount of 6.022 ×10(index 23)
∴12g/6.02210(index 23) ×1.6g
Answer : The concentration of NOBr after 95 s is, 0.013 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant =
t = time taken = 95 s
[A] = concentration of substance after time 't' = ?
= Initial concentration = 0.86 M
Now put all the given values in above equation, we get:
![0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)](https://tex.z-dn.net/?f=0.80%3D%5Cfrac%7B1%7D%7B95%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.86%29%7D%5Cright%29)
[A] = 0.013 M
Hence, the concentration of NOBr after 95 s is, 0.013 M
Molar mass CH4 = 16.0 g/mol
* number of moles:
932.3 / 16 => 58.26875 moles
T = 136.2 K
V = 0.560 L
P = ?
R = 0.082
Use the clapeyron equation:
P x V = n x R x T
P x 0.560 = 58.26875 x 0.082 x 136.2
P x 0.560 = 650.76
P = 650.76 / 0.560
P = 1162.07 atm
Answer:
It is mentioned that the student is mixing chemicals A and B and observes the time taken for the color to change. However, in the experiment, it is noticed that the student has repeated the procedure five times and each time he or she is modifying the concentration of chemical B. Thus, it is clear that the concentration of chemical B is the independent variable in the experiment. An independent variable is illustrated as the variable, which is controlled or modified in the experiment.