A few drops of a mixture of sodium hydroxide(NaOH) solution and copper (II) tetraoxosulphate (VI) solution were added to a sampl
e of urine in a test tube.State the two possible results of the test.What corresponding inferences can be made from the results above?
1 answer:
Answer:
It is a test for proteins in urine
Explanation:
The Biuret reagent is made of sodium hydroxide (NaOH) and hydrated copper(II) sulfate. The biuret reagent is commonly used to test for proteins. The biuret test is also known as Piotrowski's test. It is a chemical test commonly used in detecting the presence of peptide bonds. In the presence of peptides, a copper(II) ion forms purple-colored coordination complexes in an alkaline solution.
Hence the addition of this Biuret reagent to a urine sample in a test tube aims to detect the presence or absence of proteins in the given urine sample. If there is protein in the urine sample, the blue colour of the Biuret reagent turns purple. If there is no protein in the urine sample, the Biuret reagent remains blue.
In an alkaline solution, copper II is able to form a complex with the peptide bonds in proteins. Once this complex has been formed, the Biuret solution turns from a blue color to a purple color. This is the positive test for proteins.
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Answer: (a) cesium and bromine :
(b) sulfur and barium:
(c) calcium and fluorine :
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
The nomenclature of ionic compounds is given by:
1. Positive is written first.
2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
(a) cesium and bromine: Here cesium is having an oxidation state of +1 called as cation and bromine
is an anion with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral
called as cesium bromide.
(b) sulfur and barium : Here barium is having an oxidation state of +2 called as cation and sulphur
is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral
. called as barium sulfide.
(c) calcium and fluorine: Here calcium is having an oxidation state of +2 called as cation and fluorine
is an anion with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral
. called as calcium fluoride.
Answer:
Explanation:
Multiply by t on both sides.
Now divide by I to isolate t.