Answer:82.7% C and 17.3% H
Explanation:
C4 = 12.0107g/mol x 4 = 48.0428 g/mol
H10 = 1.00794 g/mol x 10 = 10.0794 g/mol
Molar Mass of Butane, C4H10 = 48.0428 g/mol + 10.0794 g/mol= 58.1222 g/mol
percent composition of Carbon is = ( mass of carbon contained in butane / molar mass of Butane) x 100
=(48.0428 /58.1222) x 100% = 0.8265 x 100
=82.65% =82.7% of Carbon.
Percent composition of Hydrogen = (mass of Hydrogen contained in Butane / molar mass of Butane )x 100
( 10.0794/58.122 ) x 100% =0.1734 x 100
= 17.3% OF Hydrogen.
We have to get the stable atom formed after positron emission from Terbium-147.
The stable atom is (D) ₆₄Gd¹⁴⁷.
Positron is radioactive decay. Positron is a type of beta particle β⁺.
Positron emission decreases proton number relative to neutron number, positron decay results in nuclear transmutation, changing an atom of one chemical element with an atomic number that is less by one.
Terbium on positron emission produces Gadolinium with one atomic number less than Terbium. So, the positron emission reaction is as shown below:
Tb¹⁴⁷→ ₆₄Gd¹⁴⁷ + ₁e⁰
Answer: 2) Chloroform & Caustic potash
Explanation:
The carbylamine reaction is a kind of chemical test which is done to detect primary amines in an unknown solution. It cannot detect secondary and tertiary amines.
The reaction involves the heating with up of the unknown solution with alcoholic potassium hydroxide or caustic potash and the chloroform.
In the presence of primary amine, the production of isocyanide results.
Answer:
This tells us that for every 2 moles of potassium chlorate that are decomposed, then 3 moles of oxygen gas is produced.
Mutualism! The tree provides a home and food for the ants, while the ants protect the tree :)
