I believe it is B. I’m sorry if i’m wrong. Tell me if i’m wrong or right!
0.250 mol/L
<em>Step 1</em>. Write the chemical equation
H2SO4 + 2NaOH → Na2SO4 + 2H2O
<em>Step 2</em>. Calculate the moles of H2SO4
Moles of H2SO4 = 12.5 mL H2SO4 × (0.500 mmol H2SO4/1 mL H2SO4)
= 6.25 mmol H2SO4
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 6.25 mmol H2SO4 × (2 mmol NaOH/(1 mmol H2SO4)
= 12.5 mmol NaOH
<em>Step 4</em>. Calculate the concentration of the NaOH
[NaOH] = moles/litres = 12.5 mmol/50.0 mL = 0.250 mol/L
The Correct answer to this question is translation
Ag - 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d⁹
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.