Here we have to get the product between the reaction of butane-1-amine with methyl iodide (CH₃I).
The reaction between 1 mole of butan-1-amine and 1 mole of methyl iodide produces Methyl-butamine which is a secondary amine.
However, In presence of 2 moles of methyl iodide the reaction proceed to N, N-di-methylbutamine. The reaction is shown in the figure.
This is one of the effective reaction method to generate secondary and tertiary amine from primary amine.
The primary amine reacts with alkyl iodide to form secondary to tertiary amine. The final product depends upon the quantity of the alkyl iodide present in the reaction.
The new volume : 21.85 ml
<h3>Further explanation</h3>Given
V1=25,0 ml
P1=725 mmHg
T1=298K is converted to
T2=273'K
P2=760 mmHg atm
Required
V2
Solution
Combined gas law :
Input the value :
V2=(P1.V1.T2)/(P2.T1)
V2=(725 x 25 ml x 273)/(760 x 298)
V2=21.85 ml
Answer:
510 g NO₂
General Formulas and Concepts:
Explanation:
<u>Step 1: Define</u>
6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)
<u>Step 2: Define conversions</u>
Avogadro's Number
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol
<u>Step 3: Use Dimensional Analysis</u>
<u /> = 511.901 g NO₂
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules.</em>
511.901 g NO₂ ≈ 510 g NO₂