The five main branches of chemistry are organic chemistry, inorganic chemistry, analytical chemistry, physical chemistry and biochemistry. Chemistry can be further divided into many sub-branches that may fall under more than one of the main branches.
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer:
pH = 2.66
Explanation:
- Acetic Acid + NaOH → Sodium Acetate + H₂O
First we <u>calculate the number of moles of each reactant</u>, using the <em>given volumes and concentrations</em>:
- 0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid
- 1.0 M NaOH * 10.0 mL = 10 mmol NaOH
We<u> calculate how many acetic acid moles remain after the reaction</u>:
- 37.5 mmol - 10 mmol = 27.5 mmol acetic acid
We now <u>calculate the molar concentration of acetic acid after the reaction</u>:
27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M
Then we <u>calculate [H⁺]</u>, using the<em> following formula for weak acid solutions</em>:
- [H⁺] =

Finally we <u>calculate the pH</u>: