Amines are derivatives of
Ammonia (NH₃) in which atleast one hydrogen atom is replaced by an alkyl group. Amines are further classifies as;
Primary Amines: In primary amines the nitrogen atom is attached to two hydrogen atoms and one alkyl group.
Secondary Amines: In secondary amines the nitrogen atom is attached to two alkyl groups and one hydrogen atom.
Tertiary Amines: In tertiary amines the nitrogen atom is attached to three alkyl groups, hence it has no hydrogen atom.
Below are three isomers of tertiary amines with molecular formula
C₅H₁₃N.
Answer:
Caesium (55Cs) has 40 known isotopes, making it, along with barium and mercury, one of the elements with the most isotopes. The atomic masses of these isotopes range from 112 to 151. Only one isotope, 133Cs, is stable. The longest-lived radioisotopes are 135Cs with a half-life of 2.3 million years, ... It constitutes most of the radioactivity still left from the Chernobyl accident ...
48.1 grams.
<h3>Explanation</h3>
How many moles of the reaction is needed to provide that 335 kJ of heat?
- ΔH = -824.2 kJ / mol according to the equation. In other words, each mole of the reaction releases 824.2 kJ of heat. (The value of ΔH is negative, which means that the reaction produces energy.)
- 335 / 842.2 = 0.4307 mol. It takes 0.4307 moles of the reaction <em>shown in the equation</em> to produce all the 335 kJ of heat.
How many moles of iron, Fe in 0.4307 moles of the reaction?
- The coefficient in front of Fe in the equation is two. As a result, each mole of this reaction would consume two moles of Fe. The 0.4307 moles of the reaction will consume 2 × 0.4307 = 0.861 moles of Fe to produce 335 kJ of heat.
What's the mass of 0.8614 moles of iron, Fe?
- Refer to a modern periodic table, the relative atomic mass of iron Fe is 55.85. As a result, the mass of one mole of iron Fe is 55.85 grams.
- 0.861 × 55.85 = 48.1 grams. Therefore, it takes 48.1 grams of iron to provide 335 kJ of heat through this reaction.
Answer:
1019.27 g.
Explanation:
- For the balanced reaction:
<em>2Na + Cl₂ → 2NaCl,</em>
It is clear 2 moles of Na with 1 mole of Cl₂ to produce 2 moles NaCl.
- Firstly, we need to calculate the no. of moles of Cl₂ is needed to react with 57.5 mol Na:
2 moles of Na need → 1 mol of Cl₂, from the stichiometry.
57.5 moles of Na need → ??? mol of Cl₂.
<em>∴ The no. of moles of Cl₂ is needed to react with 57.5 mol Na =</em> (1 mol)(57.5 mol)/((2 mol) <em>= 28.75 mol.</em>
<em>∴ the mass of Cl₂ is needed to react with 57.5 mol Na = (no. of moles of Cl₂)(molar mass of Cl₂) =</em> (28.75 mol)(35.453 g/mol) <em>= 1019.27 g.</em>