Answer:
The cyanidin indicator turns blue within a pH range of 5 - 7. The pH of the solution could be 5, 6 or 7.
An indicator is used to determine the endpoint of a titration.
Explanation:
Cyanidin indicator changes colour with each change in pH. In acidic solutions (pH < 7) cyanidin indicator will turn red, through to purple and blue, while in basic solutions (pH > 7), cyanidin indicator will change colour from aquamarine through to green and yellow. The cyanidin indicator turns blue within a pH range of 5 - 7.
Titration is a technique used in analytical chemistry to determine the unknown concentration of a solution. A solution of known concentration is added from a burette to the solution of unknown concentration until the reaction between the two solutions is complete. This known as the endpoint of the experiment. The endpoint of a titration is determined using an indicator which is added to reaction mixture. A colour charge is produced by the indicator at the endpoint of the reaction.
Note: An indicator is a dye of weak organic acids or bases which changes colour with changes in the pH of a solution. Some common indicators are methyl orange, methyl red, phenolphthalein, etc. These indicators are used to monitor the changes in the pH of solutions during a reaction.
Answer:
In a series circuit, all components are connected end-to-end, forming a single path for current flow. In a parallel circuit, all components are connected across each other, forming exactly two sets of electrically common points.
Explanation:
Answer:
a) Mass of solid W crystallized out 0°C is 141.564 g
b) 91.92% is the percent recovery.
Explanation:
The solubility of solid W in water at 100°C = 21.3 g/100 mL
Volume of water at 100C required to dissolve 154.0 g of W: x
723.0 mL of water will dissolve 154.0 grams of solid W.
The solubility of solid W in water at 0°C = 1.72 g/100 mL
Mass of W soluble in 723.00 mL of water at 0°C = m
a) Mass of solid W crystallized out 0°C : 154.0 g- 12.436 g=141.564 g
b) Percent recovery of solid at 0°C:
91.92% is the percent recovery.
Answer is: 4 atoms of nitrogen.
One formula unit of magnesium nitrate (Mg(NO₃)₂) has two atoms of nitrogen, so two formula units of magnesium nitrate will have four atoms of nitrogen (2·2=4).
Oxidation number of magnesium in magnesium nitrate is +2 and oxidation number of nitrate anion is -1 (2·(-1) = -2).
