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.
Answer:
35.8 g
Explanation:
Step 1: Given data
Mass of water: 63.5 g
Step 2: Calculate how many grams of KCl can be dissolved in 63.5. g of water at 80 °C
Solubility is the maximum amount of solute that can be dissolved in 100 g of solute at a specified temperature. The solubility of KCl at 80 °C is 56.3 g%g, that is, we can dissolve up to 56.3 g of KCl in 100 g of water.
63.5 g Water × 56.3 g KCl/100 g Water = 35.8 g KCl
C. Answer correct I think…
Answer:
strontium oxide + hydrochloric acid (HCL) = strontium chloride
Answer:
Explanation:
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