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Kryger [21]
3 years ago
12

What is a control used for in an experiment?

Chemistry
1 answer:
Setler79 [48]3 years ago
6 0
The control is what you keep the same for each one. For example if you are doing an experiment with plants say you water one with lemon juice, one with milk and one with warm water. The control would be they all have the same amount of sun light and soil. So the growing is equal to keep the experiment fair.
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This is the question pls help me
Kryger [21]

Answer:  B

Explanation:  Graph B compares the two temperatures on separate lines so that we can see the comoparison directly, as a function of time.  Not only does the graph quickly answer which condition id most favorable to colony growth, but it also hints at some behaviors that may accelerate growth as time goes on.  Graph C is a possible answer, if the <u>only</u> question is which promotes growth the fastest.   But the questions asks "compare," which Graph B does not allow as well as Graph C.

4 0
3 years ago
Help ASAP please. I don’t understand.
WARRIOR [948]

Answer:

a

Explanation:

4 0
3 years ago
Predict the energy trend as a substance changes from a solid to liquid to gas on this graph.
Vladimir [108]
Answer:
            Option-A (<span> It would increase from bottom left to top right) is the correct answer.

Explanation:
                   As we know converting solids into liquids and converting liquids into gases require energy. This energy provided increases the energy of the state and its particles start moving with higher velocities. Therefore, the energy of solids will be lower than liquids and gases respectively. While, liquids have greater energy than solids but less energy than gases. And, gases are the most energetic than solids and liquids.</span>
8 0
3 years ago
Read 2 more answers
If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of T
Simora [160]

Answer:

The new pH after adding HCl is 7.07

Explanation:

The formula for calculating pH of a buffer is

pH = pKa + log([Conjugate base]/[Acid])

<u>Before adding HCl,</u>

         7.55 = 7.55 + log([Conjugate base]/[Acid])

⇔      log([Conjugate base]/[Acid])  = 0

⇔     [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M

⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol

<u>After adding HCl (3.00 mL, 1.00 M)</u>

⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)

New volume solution is 120 m L+ 3 mL = 123 mL

HCl is a strong acid, it will convert the conjugate base to acid form, or we can express

Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol

                ⇒ Concentration = 0.003/0.123 M

Mole of new Acid form = 0.006 + 0.003 = 0.009 mol

                ⇒ Concentration = 0.009/0.123 M

Use the formula

pH = pKa + log([Conjugate base]/[Acid])

    = 7.55 + log(0.003 / 0.009) = 7.07

8 0
3 years ago
Your friend challenged you to a race you know in order to beat him you must run 15 meters within 20 seconds in the northern dire
Marat540 [252]

Answer:

.75 meter north

Explanation:

v = d/t

v = 15/20

v = .75

7 0
3 years ago
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