Answer:
Two, KCl and PbCl₂.
Explanation:
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In this case, since the addition of chloride ions promote the following three ionic reactions:
We can infer that both silver chloride and lead (II) chloride are precipitated products as their Ksp are 6.56x10⁻⁴ and 1.59x10⁻⁵ respectively, which means they are merely soluble in water.
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Answer:
The 2s orbital is at a higher energy level.
Explanation:
1s and 2s are the sub-orbitals that are located in an atom. They are nearest to the nucleus and are found on the s sub-orbital. The difference between 1s and 2s is the difference in their level of energy. 1s has low energy as compared to 2s. 1s orbital has the lowest energy because it is located closed to the nucleus. 2s orbital has higher energy than 1s because it's orbit is larger than 1s.
Answer;
-(2) An atom is mostly empty space.
Experiment
-Rutherford conducted the "gold foil" experiment where he shot alpha particles at a thin sheet of gold. The conclusion that can be drawn from these experiment is that an atom is mostly empty space.
-Rutherford found that a small percentage of the particles were deflected, while a majority passed through the sheet. This caused Rutherford to conclude that the mass of an atom was concentrated at its center, as the tiny, dense nucleus was causing the deflections.
The partial pressure is the amount of linguistic compound there is which makes the lagitude of the element 64.663
Answer:
5.37 × 10⁻⁴ mol/L
Explanation:
<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.00154 mol/L
- Initial volume (V₁): 230. mL
- Final concentration (C₂): ?
- Final volume (V₂): 660. mL
Step 2: Calculate the concentration of the final solution
We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁ / V₂
C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L
