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USPshnik [31]
3 years ago
10

What is the purpose of the scientific method? ON APEX LEARNING

Chemistry
2 answers:
Arturiano [62]3 years ago
4 0

The purpose of the scientific method "to use an experiment to test a scientific theory".

<u>Answer:</u> Option C

<u>Explanation:</u>

Researchers use the scientific method when conducting experiments to collect observable, empirical data in a hypothesis-related experiment , the findings that tend to justify or oppose a theory. It's also very important that other researchers can repeat and independently verify the conclusions, logic, and experiments.

Scientific method retains inductive reasoning its position. Scientists use this to formulate theories and hypotheses. In order to adapt the ideas to particular situations, deductive reasoning allows them to do so. Overall scientific methods have become path of analyzing different processes, going in our environment.

WITCHER [35]3 years ago
3 0

Answer:

answer is b not c

Explanation:

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If you were asked to convert 25 cg to the unit hg, which of the following would be the first fraction used in the conversion? on
zhannawk [14.2K]
We are asked to convert 25 cg to units of hg.

1 cg = 1 centigram = 10⁻² g
1 hg = 1 hectogram = 10² g

The options given are:

a) 1 hg/ 10² g
b) 10² cg/ 1 hg
c) 10² hg/ 1 cg
d) 10⁻² g/ 1 cg

To convert 25 cg to 1 hg, we could convert the 25 cg to grams first, then grams to hg.

25 cg · 10⁻² g/ 1cg = 0.25 g

Here we have converted our number from cg to grams. We can use another conversion of grams to hg to complete the conversion.

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3 years ago
In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O
yarga [219]
HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:

HClO + H2O ⇄ H3O+ +OCl-

Then the equilibrium constant, Ka, of dilute HClO would be:

K_{a} = \frac{[ H_{3}  O^{+} ][O Cl^{-} ]}{HClO}

Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
      
          HClO       + H2O   ⇄   H3O+ +  OCl-
I     1.0x10^-4                          0             0
C        -x                                 +x           +x 
E  (1.0x10^-4 - x)                     x             x

Substituting the excess (E) concentration to the Ka equation:

K_{a} = \frac{[x ][x]}{1.0 \ x \  10^{-4} - x }

Simplifying the equation would yield a quadratic equation:

x^{2} + K_{a}x-(1.0 \ x \ 10^{-4}) K_{a}=0

The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.

x = 1.688 x 10^-6 M = [H3O+] = [ClO-]

Then, you can determine the conc of [OH-] through pH.

pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M

Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M

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[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.

Thus in relative amounts, the order would be

H2O >>> HClO > H3O+ = ClO- > OH-


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