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Find the concentration of benzene in water (miligram/Liter) if the water is in equilibrium with a layer of oil containing 1%(mol) benzene. the solubility of benzene in water is 1.8 gram/Liter, the molecular weight of benzene is 78.11g/mol
explain why there is a difference in spectrophotometric response during titration before and after the equivalence point by placing the statements below into the appropriate categories.Now, calculate the concentration of apotransferrin present in ..
Carbon disulfide normally boils at 46.5 degrees celsius. when the temperature is lowered to 28.0 degrees celcius, its vapor becomes 400 mmHG. what will be the vapor pressure if the temperature is lowered some more to 18 degrees celcius
Nitrogen has only two isotopes of appreciable natural abundance: 14N, with atomic mass of 14.00307 amu; and 15N with atomic mass of 15.00011 amu.
Write the corresponding ionic and net ionic equations after Balancing these given equations
What is the pH of a solution made by dissolving 0.052 grams of RbOH in enough water to make 64.9 mL of solution
What is the correct positions of the methyl group and the isopropyl group in the most stable chair confirmation of cis-1-isopropyl-2-methylcyclhexane
That concentration (g / ml) should be prepared with a substance weight of 200g/mol melocular to give an absorbance of 0.8? Assume that the substance has a molar absorbance coefficient of 16,000 and that the cell has a light path length of 1 cm.
Find the pH during the titration of 20.00 mL of .1000 M butanoic acid CH3CH2CH2COOH (kA = 1.54E-5), with .1000 M NaOh solution after the following additions of titrant. a. 20.40 mL b. 26.00 mL
rewrite the equation so heat of reaction (delta H) equals deltaHf for 4PCl3 (g) yields p4(s) + 6Cl2 (g) Values for delta HF in Kj/mol are:
there are some solutions that are given as a. ch4nbspb. caoh2nbspc. naohnbsp d. co2nbsp e. ch3nh2 find out which
1) Bakelite us obtained by the condensation of : 2) When HCHO reacts with Nh3, the compound formed is :
What is the pH of the solution created by combining 2.00 mL of the 0.10 M base with 8.00 mL of the 0.10 M acid
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