Answer:
B and D could be true
Explanation:
A volume of sodium hydroxide less than expected could occurs for two reasons:
The real concentration of sodium hydroxide was higher than expected or the amount of vinegar added was less than expected:
A. The sodium hydroxide solution had been allowed to stand exposed to the air for a long time prior to the titration. FALSE. A long expose to the air decreases concentration of the NaOH.
B. The volumetric flask used to prepare the diluted vinegar solution was rinsed with water prior to use. TRUE. You add a less amount of vinegar doing you require less amount of NaOH than expected.
C. The burette used to deliver the sodium hydroxide solution was rinsed with water prior to use. FALSE. Thus, you add a less amount of NaOH than expected. To explain the matter, you add more NaOH than expected.
D. The pipette used to deliver the vinegar solution was rinsed with water prior to use. TRUE. Again, you are adding a less amount of Vinegar than expected doing the necessary NaOH during titration less than expected
Answer:
C6H12O6 + 6O2 → 6CO2 + 6H2O.
Explanation:
The temperature of something.
Answer:
148 g
Explanation:
Step 1: Write the balanced equation for the decomposition of sodium azide
2 NaN₃ ⇒ 2 Na + 3 N₂
Step 2: Calculate the moles corresponding to 95.8 g of N₂
The molar mass of N₂ is 28.01 g/mol.
95.8 g × 1 mol/28.01 g = 3.42 mol
Step 3: Calculate the moles of NaN₃ needed to form 3.42 moles of N₂
The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.42 mol = 2.28 mol.
Step 4: Calculate the mass corresponding to 2.28 moles of NaN₃
The molar mass of NaN₃ is 65.01 g/mol.
2.28 mol × 65.01 g/mol = 148 g
