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Sonbull [250]
3 years ago
5

What does it mean for a hypothesis to be testable?

Chemistry
2 answers:
KengaRu [80]3 years ago
8 0
It means that you can test the question that you are asking for the topic or particular project that you are focusing on
Gwar [14]3 years ago
6 0
It means that it has to be able to be test in an expermint that your doing
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Due Tuesday 09-07-2021 Please help immediately!! I have 10 missing assignments from being absent
jeka57 [31]

Answer:

Go to

On the left, click Missing.

(Optional) To review more details, click an item. View details.

Explanation:

7 0
2 years ago
Read 2 more answers
Lithium and fluorine undergo ionic bonding. Using the noble gas electron configurations for each (below), please explain the pro
Slav-nsk [51]

Answer:

Lithium loses one electron to fluorine and forms ionic bond, having formula LiF.

Explanation:

Lithium is the element of the group 1 and period 2 which means that the valence electronic configuration is [He]2s^1.

Fluorine is the element of the group 17 and period 2 which means that the valence electronic configuration is [He]2s^22p^5.

<u>Thus, lithium loses 1 electron and become positively charged. Fluorine on the other hand accepts this electron and become negatively charged.</u> This is done in order that the octet of the atoms are complete.  <u>These both ions then form ionic bond as their will be electrostatic interaction between the two oppositely charged ions.</u>

Thus, the formula of calcium chloride is LiF.

7 0
3 years ago
Which property is characterized by the ability to bend?
galben [10]
Answer: malleable

explanation: malleable means to be easily manipulated in shape
6 0
3 years ago
Read 2 more answers
What is the hybridization and shape for an XeF6 2+ molecule?
Anna11 [10]

Answer:

Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²

Explanation:

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²

Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48

Ca. #Substrate e⁻ = 6F = 6(8) = 48

#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs

#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs

BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²    

XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)      

8 0
3 years ago
Consider the reaction: A &lt;=&gt; B. Under standard conditions at equiliubrium, the concentrations of the compounds are [A] = 1
Katen [24]

Answer:

Keq'>1\\\Delta G'

Explanation:

Hello,

In this case, for the given reaction, the equilibrium constant turns out:

Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3

Nonetheless, we are asked for the reverse equilibrium constant that is:

Keq'=\frac{1}{Keq}=3

Which is greater than one.

In such a way, the Gibbs free energy turns out:

\Delta G'=-RTln(Keq')\\

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.

7 0
3 years ago
Read 2 more answers
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