Answer:
3. turns pink and is basic
Explanation:
Phenolphthalein is a pH indicator that remains colorless in acidic solutions, but in basic solutions it turns pink at a pH equal to 10.
Phenolphthalein is a weak acid that loses H+ cations in solution. The phenolphthalein molecule is colorless, while the phenolphthalein-derived anion is pink. When a base is added, phenolphthalein loses H+, forming the anion and causing it to turn pink. The color change cannot be explained only on the basis of deprotonation, a structural change occurs with the appearance of a ketoenolic tautomerism.
Answer:
Forensic drug chemists analyze samples of unknown materials including powders, liquids and stains to determine the chemical identity or characteristics of the compounds that make up the sample. samples submitted as evidence in a drug-related case can contain one compound or a mixture of many compounds.
First, we apply the law of conservation of mass which states that the total mass in a system remains constant.
Therefore, there must be 5.00 g of sulfur and 4.99 g of oxygen in the product. Now, we determine the mass percentage using:
Mass % = (mass of sulfur x 100) / total mass of compound
Mass % = (5 * 100) / (5 + 4.99)
Mass % = 50.05%
The product contains 50.05% sulfur by mass.
1) Chemical reaction: AgNO₃ + HCl → AgCl + HNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(HCl) = 1 : 1.
0,00675 mol : n(HCl) = 1 : 1.
n(HCl) = 0,00675 mol.
V(HCl) = n(HCl) ÷ c(HCl).
V(HCl) = 0,00675 mol ÷ 0,130 mol/L.
V(HCl) = 0,0519 L = 51,92 ml.
2) 1) Chemical reaction: AgNO₃ + KCl → AgCl + KNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(KCl) = 1 : 1.
0,00675 mol : n(KCl) = 1 : 1.
n(KCl) = 0,00675 mol.
m(KCl) = n(KCl) · M(KCl).
m(KCl) = 0,00675 mol · 74,55 g/mol.
m(KCl) = 0,503 g.
n - amount of substance.
M - molar mass.
Explanation:
2H2(g) + O2(g) → 2H2O(l )