Valence electrons are the electrons in the outermost shell. Those are t<span>he electrons on an atom that can be gained or lost in a chemical reaction.
</span>Elements that are left on the periodic table <span> have relatively few </span>valence electrons<span>, and can form ions more easily by losing their </span>valence electrons<span> to form positively charged cations.</span>
<span>Nonmetals are further to the right on the periodic table, so they gain electrons relatively easily and lose them with difficulty. </span>
I. The solubility of NaCl at 25 degrees C would be between the solubilities at 20 and 30 degrees C. A reasonable answer would be 36 grams/100 g water
ii. From the table, it’s clear that the salts are more soluble at higher temperatures, indicating that an increase in temperature increases solubility.
iii. At 50 degrees C, a saturated ammonium chloride solution will have 50.6 grams of salt per 100 g water. At 20 degrees C, the solution can hold only 37.3 grams of salt per 100 g water. Thus, 13.3 grams of salt will precipitate per 100 grams of water.
The epicenter was located somewhere on a circle centered at Recording station X, with a radius of 250 km.<span>
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You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.
The total number of ions in 38.1 g of SrF₂ is 5.479 x 10²³.
<h3>What are ions?</h3>
Ions are the elements with a charge on them. It happens when they share electrons with other atoms to form a compound.
We have to calculate the total number of ions in 38.1 g of .
The molar mass of SrF₂ = 125.62 g/mol
The number of moles = 38.1 g of 1.0 mol / 125.62 = 0.30329 moles
Given that, total moles of SrF₂ ions in = 1.0 mol of + 2.0 moles of = 3.0 moles
Total moles of ions in 0.30329 moles of
= (0.30329 moles of SrF₂) x 3.0 / 1.0 = 0.90988 mol ions
We know that,
1.0 mole of ions = 6.023 x 10²³ ions
Thus, the number of total ions = ( 0.90988 mol ions) x 6.023 x 10²³ / 1.0 mol = 5.479 x 10²³ ions
Thus, the number of ions is in 38.1 g of 5.479 x 10²³ ions
To learn more about ions, refer to the link:
brainly.com/question/14295820
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