If your spectrophotometer can measure an absorbance up to 1.5 what is the maximum concentration

It depends of the experiment

Explanation:

The measurement of the concentration depends on the way in which the quantification is being carried out.

The first method is based on the law of lamber-beer, A = bce. Where a is the absorbance, b is the optical path and is the molar extension coefficient. All these variables will depend on the method that is being carried out. For example, when you want to measure immunogluboline, you have a molar extinction coefficient of 210,000M-1 cm-1, the optical path is always 1 cm and for this case the absorbance is 1.5, therefore the concentration will be:

The second procedure is the calibration curve method. In this case, we have to measure the concentration of several samples and register the absorbance for each sample. In this procedure, we will obtain a lineal formula:

Where Y is is the absorbance, m is the slope, X is the concentration and b the intercept with the Y-axis. For example, when we messure the concentration of we will have the concentrations (figure 1).

The curve obtained through a linear regression will be:

If we replace the value in the equation we will obtain:

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied