Answer:
0.1 g/dl
Explanation:
The standard curve is a graph that relates the absorbance at 400 nm with the concentration of haemoglobin in mg/dl. To obtain the concentration from the absorbance value, we enter in the x-axis (absorbance at 400 nm) with the value 0.40 (the line between 0.2 and 0.6), we extrapolate the line to the curve and read the correspondent value on y-axis (concentration in mg/dl): 100 mg/dl.
So, we convert the concentration from mg/dl to g/dl by dividing into 1000:
100 mg/dl x 1 g/1000 mg = 0.1 g/dl
Therefore, the concentration of haemoglobin of the patient is 0.1 g/dl.
Answer:
A)The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.
B)Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.
C)If the product is wet with water there will be no change in the infrared spectrum
Explanation:
The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.
Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.
If the product is wet with water there will be no change in the infrared spectrum
Answer:
No.
Explanation:
No. There is 1 atom of Ca on the left and 2 Ca's on the right and 2 OH's on the left and 4 on the right.
The balanced equation is:
4OH- + 2Ca2+ ----> 2Ca(OH)2.
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of 
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration .
The given equilibrium reaction is,
Initially c 0
At equilibrium 
The expression of 
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:
Therefore, the value of equilibrium constant for this reaction is, 1.1
