Answer:
0.0693M Fe
Explanation:
It is possible to quantify Fe in a sample using Mn as internal standard using response factor formula:
F = A(analyte)×C(std) / A(std)×C(analyte) <em>(1)</em>
Where A is area of analyte and std, and C is concentration.
Replacing with first values:
F = 1.05×2.00mg/mL / 1.00×2.50mg/mL
<em>F = 0.84</em>
In the unknown solution, concentration of Mn is:
13.5mg/mL × (1.00mL/6.00mL) = <em>2.25 mg Mn/mL</em>
Replacing in (1) with absorbances values and F value:
0.84 = 0.185×2.25mg/mL / 0.128×C(analyte)
C(analyte) = <em>3.87 mg Fe / mL</em>
As molarity is moles of solute (Fe) per liter of solution:
= <em>0.0693M Fe</em>
Answer:
<h3>F=4k.gm/s^2</h3>
Explanation:
<h3>F=m×a</h3><h3>f=2k.g×2m/s^2</h3><h3>f=4k.gm/s^2</h3>
Answer:
Chemical solution in solid form; whose solvent's crystal structure is not altered by solute
Explanation:
Answer:
thats all help me too:)...............
Answer : The rate constant at 785.0 K is, 
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= ?
= activation energy for the reaction = 262 kJ/mole = 262000 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature =
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B6.1%5Ctimes%2010%5E%7B-8%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B262000J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B785.0K%7D%5D)
Therefore, the rate constant at 785.0 K is,
