Answer:
2.17 gm Al
Explanation:
First, find the number of moles . Then multiply by the mole weight of aluminum.
# moles = 4.85 x 10^22 / 6.022 x 10^23 = .08053 moles
.0805 moles * 26.981 gm/mole = 2.173 gm
Answer:
Weak acid
Explanation:
A titration curve is a graphical description of the change in pH of the solution in the conical flask as the reagent is added from the burette. A titration curve can be plotted for the different kinds of acid and base titrations. The volume of the titrant is always plotted as the independent variable and the pH of the solution as the dependent variable. The equivalence point is read off from the titration curve. A titration curve is very important because it shows the pH at various points during the titration.
A weak acid/strong base titration leads to an equivalence point above 7. From the question, we were told that the pH at equivalence point lies around 8. Hence the unknown substance must be a weak acid.
Answer : Option B) All atoms of a single substance are identical.
Explanation : The scientist John Dalton proposed the atomic theory which had the postulates as follows.
i) All matter/substances consists of indivisible particles known as atoms.
ii) Atoms of the same element/substance are similar in mass,shape and size, but differ from the atoms of other elements.
iii) Atoms obey the law of conservation of energy which says atoms cannot be created or destroyed.
iv) Atoms of different elements may combine with each other in a fixed, simple, whole number ratios to form any compound atoms.
v) Atoms of same element can combine in ratio with more than one to form two or more compounds.
vi) The atom is considered to be the smallest unit of matter that can take part in a chemical reaction
Students performed a procedure similar to Part II of this
experiment (Analyzing Juices for Vitamin C Content) as described in the
procedure section. Given that molarity of DCP is 9.98x10-4 M, it took 16.34 ml
of DCP to titrate 10 mL of sample.
Amount of ascorbic acid = 0.050 L sample (0.01634 L DCP/0.01
L sample)( 9.98x10-4 mol DCP/L DCP)(1 mol Ascorbic acid/ 1mol DCP)(176.124
g/mol)(1000mg/1g)= 14.36 mg ascorbic acid
Answer:
IR spectroscopy can be used to identify chemical structures are present in compounds.
Explanation:
Infrared spectroscopy is a technique in organic chemistry that can be use use to identify chemical structures present in compounds because it is base on the ability of different functional groups to adsorb infrared light.
This work by shinning the infrared lights into the organic compounds to be identified, some of the frequencies of the infrared lights are adsorbed by the compounds and its identify groups of atoms and molecules in the compound.
