Answer:
we do the best possible in our
Answer:
For this experiment we are going to take plate 1 as the control plate, so, in it there will be just E. coli in LB/agar; in plate 2, we are going to put E. coli in LB/agar and some ampicillin. Then, we have to wait for the E. coli colonies to form. After a while, the E. coli growth can be compared on both plates and determine if ampicillin affects or not the E. coli colonies.
Explanation:
If the ampicillin affects negatively E. coli colonies, we are going to observe that in plate 1 (control plate) there are E. coli colonies growing, but in plate 2, there is no E. coli colonies or, at least, there is a fewer number of colonies on it. If ampicillin doesn't affect E.coli, plate 1 (control) and plate 2 (ampicillin experiment) are going to be similar in number of colonies.
Answer:
strong acid because it dissociates almost completely.
Answer:
The specific heat of the sample unknown metal is approximately 0.45 J/g °C.
General Formulas and Concepts:
<u>Thermodynamics</u>
Specific Heat Formula: 
- <em>m</em> is mass (g)
- <em>c</em> is specific heat capacity (J/g °C)
- Δ<em>T</em> is the change in temperature
Explanation:
<u>Step 1: Define</u>
<em>Identify variables.</em>
<em>m</em> = 112 g
Δ<em>T</em> = 20.0 °C
<em>q</em> = 1004 J
<u>Step 2: Solve for </u><u><em>c</em></u>
- Substitute in variables [Specific Heat Formula]:

- Simplify:

- Isolate <em>c</em>:

- Round [Sig Figs]:

∴ specific heat capacity <em>c</em> is equal to around 0.45 J/g °C.
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Topic: AP Chemistry
Unit: Thermodynamics