Answer:
6.25 μg/mL
Explanation:
When a dilution is made, the mass of the solute is conserved (Lavoiser's law), so the mass pipetted will be the mass in the assay. The mass is the concentration (C) multiplied by the volume (V). If the pipet solution is called 1, and the assay 2:
m1 = m2
C1*V1 = C2*V2
C1 = 250 μg/mL
V1 = 25 μL
V2 = 975 μL + 25 μL = 1000 μL (is the final volume of the assay after the addition of LDH)
250*25 = C2*1000
C2 = 6.25 μg/mL
Answer:
here
Explanation:
0.000141 to kilowatt-hours. hope this helped
Solve first for the number of moles of sodium given that the mass is 2500 g by dividing the given mass by the molar mass of sodium.
moles sodium = 2500 g / 23 g/mol
moles sodium = 108.7 moles
Since in every mole of NaCl, there is only one mole of sodium then, we conclude that there are also 108.7 moles of NaCl. Multiplying the number of moles to the molar mass of sodium chloride (58.44), the answer would be 6,352.17 grams.
Energy cannot be destroyed or created, but energy could be transformed or transferred. For example a skiier skiing from the mouth can have potential energy transferred into kinetic energy.
Answer:
35Cl = 75.9 %
37Cl = 24.1 %
Explanation:
Step 1: Data given
The relative atomic mass of Chlorine = 35.45 amu
Mass of the isotopes:
35Cl = 34.96885269 amu
37Cl = 36.96590258 amu
Step 2: Calculate percentage abundance
35.45 = x*34.96885269 + y*36.96590258
x+y = 1 x = 1-y
35.45 = (1-y)*34.96885269 + y*36.96590258
35.45 = 34.96885269 - 34.96885269y +36.96590258y
0.48114731 = 1,99704989y
y = 0.241 = 24.1 %
35Cl = 34.96885269 amu = 75.9 %
37Cl = 36.96590258 amu = 24.1 %
