Pure solid NaH2PO4 is dissolved in distilled water, making 100.00 ml of solution. 10.00ml of this solution is diluted to 100.00 ml to prepare the original phosphate standard solution. Three working standard solutions are made from this by pipetting 0.8ml, 1.5ml and 3.0 ml of the original standard solution into 100.00 ml volumetric flasks. Acid and molybdate reagent are added and the solutions are diluted to 100.00 ml. You may assume that all these absorbance measurements have already been corrected for any blank absorbance. The absorbance of each is measured in the spectrophotometer. Mass of NaH2PO4 (mg) 512.2 Absorbance Standard 1 (0.8 ml) 0.1922 Standard 2 (1.5 ml) 0.3604 Standard 3 (3.0 ml) 0.7209 Calculate the following Concentration of original phosphate standard (mM) , Concentration of Standard 1(mM), Concentration of Standard 2(mM), Concentration of Standard 3(mM), Slope of calibration line?

concentration of original phosphate standard = 4.27 mM

Concentration of standard 1 = 0.0342 mM

Concentration of standard 2 = 0.0640 mM

Concentration of standard 3 = 0.128 mM

slope of calibration line = 5.63

Explanation:

mass of NaH2PO4 = 512.2 mg

moles of NaH2PO4 = (mass of NaH2PO4) / (molar mass of NaH2PO4)

moles of NaH2PO4 = (512.2 mg ) / (119.977 g/mol)

moles of NaH2PO4 = 4.27 mmol

concentration of original solution = (moles of NaH2PO4) / (volume of solution in Liter)

concentration of original solution = (4.27 mmol) / (0.10000 L)

concentration of original solution = 42.7 mM

concentration of original standard solution = (concentration of original solution) * (volume of original solution / volume of standard solution)

concentration of original standard solution = (42.7 mM) * (10.00 mL / 100.00 mL)

concentration of original standard solution = .4.27 mM

concentration of standard 1 = (concentration of original standard solution) * (volume of standard solution / volume of standard 1)

concentration of standard 1 = (4.27 mM) * (0.8 mL / 100.00 mL)

concentration of standard 1 = 0.03415 mM

concentration of standard 2 = (concentration of original standard solution) * (volume of standard solution / volume of standard 2)

concentration of standard 2 = (4.27 mM) * (1.5 mL / 100.00 mL)

concentration of standard 2 = 0.0640 mM

why do the isotopes of a particular element behave differently in nuclear reactions but the same in chemical reactions? chemical reactions involve changes to electrons in an atom; isotopes do not differ in electrons, and so they react the same way during chemical changes.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j