Answer:
repetition
Explanation:
Taking an average of 16 height measurements is an example of repetition. Repetition involves making and taking repeated measurements in an experiment.
The goal is to achieve a highly accurate and precise data from the experiment.
- Replication involves duplicating another experiment and testing to see how valid they are.
- Since Veronica's experiment is a brand new one, it's hypothesis has not been tested or replicated in any way.
- Therefore, it is not a replication.
- But she is repeating the experiment to obtain different values.
<u>Answer:</u> The rate law of the reaction is ![\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5BC_2O_4%5E%7B2-%7D%5D%5E2)
<u>Explanation:</u>
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
Rate law expression for the reaction:
![\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5Ea%5BC_2O_4%5E%7B2-%7D%5D%5Eb)
where,
a = order with respect to 
b = order with respect to 
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Expression for rate law for third observation:
....(3)
Expression for rate law for fourth observation:
....(4)
Dividing 2 from 1, we get:
Dividing 2 from 3, we get:
Thus, the rate law becomes:
![\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5E1%5BC_2O_4%5E%7B2-%7D%5D%5E2)
Answer:
390.85mL
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 780 torr
Initial volume (V1) = 400mL
Initial temperature (T1) = 40°C = 40°C + 273 = 313K
Final temperature (T2) = 25°C = 25°C + 273 = 298K
Final pressure (P2) = 1 atm = 760torr
Final volume (V2) =?
Step 2:
Determination of the final volume i.e the volume of the gas outside Matt's body.
The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:
P1V1/T1 = P2V2/T2
780 x 400/313 = 760 x V2 /298
Cross multiply to express in linear form
313 x 760 x V2 = 780 x 400 x 298
Divide both side by 313 x 760
V2 = (780 x 400 x 298) /(313 x 760)
V2 = 390.85mL
Therefore, the volume of the gas outside Matt's body is 390.85mL
