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maria [59]
3 years ago
13

Plssss heellllp!!!

Chemistry
2 answers:
mars1129 [50]3 years ago
8 0

Answer:

I feel like it is A, a theory might be supported by a scientific evidence but it might not be true overall, an opinion is someone expressing how they think or feel about something, it might also be supported by a scientific evidence but tbh I hope I got it right

Naddika [18.5K]3 years ago
7 0
A. A theory is supported by scientific evidence, while an option is not.
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6 0
2 years ago
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Answer:

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4 0
3 years ago
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How many lone pairs does CO2 have? Does it have Single bonds or Double bonds? Does it obey the octet rule?
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The equilibrium constant for the dissolution of silver chloride (AgCl(s) Ag+(aq) + Cl–(aq)) has a value of 1.79 × 10–10. Which s
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"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.

Option: b

<u>Explanation</u>:

As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant \left(\boldsymbol{K}_{s p}\right) for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures.  Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.  

To calculate \left(\boldsymbol{K}_{s p}\right) firstly molarity of ions are needed to be found with formula: \text { Molarity of ions }=\frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}

Then at equilibrium cations and anions concentration is considered same hence:

\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }

Hence from above data \left(\boldsymbol{K}_{s p}\right) can be calculated by: \left(\boldsymbol{K}_{s p}\right) = \left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]

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