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QveST [7]
2 years ago
15

How do i design a controlled experiment to appropriately test a hypothesis?

Chemistry
1 answer:
Virty [35]2 years ago
3 0

Answer:

you need only one independent variable because if not, you wont know what factors have changed your experiment.

Explanation:

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Mixtures are classified as homogeneous and heterogeneous according to _____.
Dafna1 [17]

Answer:

i believe its the 1st option

8 0
2 years ago
What happens to particles during<br> changes of state between solids,<br> liquids, and gases?
oksano4ka [1.4K]
Answer:
Their vibrations speed up

Explanation:
They start vibrating faster and faster and start generating more and more heat and separate from each other so, therefore (usually), become less dense
7 0
2 years ago
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BLOOD HELPS YOUR BODY TO STAY AT TEMPERATURE OF ABOUT _____C. *
Paladinen [302]

Answer:The average normal body temperature is most often said to be 98.6° F (37° C). ... One small study even suggested that in healthy older patients, body temperature ranged from 94° F to 99.6° F, with an average of 97.7° F. Several factors can lead to a lower body temperature in older people

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7 0
3 years ago
Arrange these elements according to atomic Pb Si C Sn Ge
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........C<Si<Ge<Sn<Pb
5 0
3 years ago
A 1.00 L buffer solution is 0.112 M in acetic acid and 0.112 M in sodium acetate. Acetic acid has a pKa of 4.74. What is the pH
topjm [15]

Answer:

ΔpH = 1.25

Explanation:

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.112] / [0.112]

<em>pH = 4.74</em>

The reaction of sodium acetate (CH₃COONa) with HCl is:

CH₃COONa + HCl → CH₃COOH + NaCl

<em>Producing acetic acid, </em>CH₃COOH.

If 0.1mol of HCl reacts the final moles of CH₃COONa are:

0.112mol - 0,1 mol = 0.012mol

Moles of acetic acid are:

0.112mol + 0,1 mol = 0.212mol

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.012] / [0.212]

<em>pH = 3.49</em>

<em></em>

Change in pH, ΔpH = 4.74 - 3.49 =<em> 1.25</em>

I hope it helps!

6 0
3 years ago
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