Question

Help with the LAB answers

Compute Reaction Rates for All Seven Trials

Reaction rate is usually computed as a change in concentration (e.g., molarity) per change in time. The tablet is not pure NaHCO3, so the molarity cannot be computed accurately. (In addition, the quantities in this lab are quite small, and the resulting values would be small decimal values.) Instead, compute the following as a measurement of reaction rate.

Answer 1

36ml, 146ml, 190ml, 352ml

Step-by-step explanation:

Answer 2

Answer : The reaction rate at
`3^oC,24^oC,40^oC\text{ and }65^oC` are
`36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec`

Solution : Given,

Mass of tablet = 1000 mg

Volume of water = 0.200 L

Formula used :

`\text{Reaction rate}=\frac{\text{Mass of tablet}/\text{Volume of water}}{\text{Reaction time}}`

Now we have to calculate the reaction rate at different temperatures and reaction time.

`\text{Reaction rate at }3^oC=(1000mg/0.200L)/(138.5sec)=(5000mg/L)/(138.5sec)=36mg/L/sec`

`\text{Reaction rate at }24^oC=(1000mg/0.200L)/(34.2sec)=(5000mg/L)/(34.2sec)=142mg/L/sec`

`\text{Reaction rate at }40^oC=(1000mg/0.200L)/(26.3sec)=(5000mg/L)/(26.3sec)=190mg/L/sec`

`\text{Reaction rate at }65^oC=(1000mg/0.200L)/(14.2sec)=(5000mg/L)/(14.2sec)=352mg/L/sec`

Therefore, the reaction rate at
`3^oC,24^oC,40^oC\text{ and }65^oC` are
`36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec`