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2 years ago

12

Ultimately, all scientific knowledge comes from A.experimentation B.observation C.textbooks D.both experimentation & observa

tion

2 answers:

serg [7]2 years ago

6 0

Answer: D

Explanation: both experimentation and observation

Ilia_Sergeevich [38]2 years ago

5 0

The answer is D, because all scientific knowledge comes from experimenting and observing things! i hope this helps! :)

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What was the key design change for hfc-134a a/c systems versus CFC 12 a/c systems

BabaBlast [244]

Answer is: key design change for HFC-134a A/C systems versus CFC-12 A/C systems was quick couple service fitting and that design reduce venting and mixing of refrigerants during service.

<span> Level of contamination is also reduced and the emission of refrigerants and greenhouse gases (sulfur dioxide, carbon dioxide) is also reduced.</span>

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3 years ago

Which of the following is an anion?<br> A. O^2- <br> C. Al^3+ <br> D. H

kifflom [539]

The correct answer is A. O^2-

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2 years ago

The total number of atoms represented by the formula k3fe(cn)6 is:

Brut [27]

<span>K3Fe(CN)6

K - 3 atoms
Fe - 1 atom
C - 6 atoms
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Altogether : 3+1+6+6= 16 atoms</span>

4 0

3 years ago

Read 2 more answers

If the reduction reaction has a reduction potential of 0.1 V, and the oxidation reaction has a reduction potential of -0.4V, and

aleksley [76]

Answer : The value of ΔG expressed in terms of F is, -1 F

Explanation :

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

or,

$E^o=E^o_{reduction}-E^o_{oxidation}$

$E^o=(0.1V)-(-0.4V)=+0.5V$

Now we have to calculate the standard cell potential.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant

$E^o$ = standard e.m.f of cell = +0.5 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times F\times 0.5)$

$\Delta G^o=-1F$

Therefore, the value of ΔG expressed in terms of F is, -1 F

5 0

3 years ago

Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super

dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below: $X^{m+}Y^{n-}\rightarrow X_nY_m$

Thus, for all the given combinations, we obtain:

- Y⁻

$X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3$

- Y²⁻

$X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3$

- Y³⁻

$X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY$

Best regards!

8 0

2 years ago