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MissTica
2 years ago
11

Refer to the image please!

Chemistry
1 answer:
Ludmilka [50]2 years ago
8 0

Answer:

9. Electron Affinity

10. The second option....

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A student who was training for a cross-country race jogged for 1.0 hour and covered a distance of 1.5 kilometers. How much farth
densk [106]

Answer:

He would travel 3 kilometers.

Explanation:

Since he had already covered 1.5 kilometers in the time of 1 hour that means if he jogged for 1 more hour assuming he is going at the same pace he would've jogged 1.5 kilometers again and when you add that to the 1.5 kilometers he started with it would add up to 3 kilometers.

4 0
3 years ago
What must group 16 do to become stable
tamaranim1 [39]
Group 16 must gain 2 electrons to become stable as stated in my other answer 

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7 0
3 years ago
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What is the % of water in (MgSO4 . 2H2O)?
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Answer: A hydrate is found to have the following percent composition: 48.8% MgSO4 and 51.2% water.

Explanation:

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7 0
3 years ago
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105
rosijanka [135]
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
6 0
3 years ago
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12 is produced by the reaction of 0.5235 mol of CuCl2 according to the following equation: 2CuCl2 +4K1—2Cul + 4KCI + 12. What is
Keith_Richards [23]

Answer:

The mass of I₂ produced is 66.43 g.

Explanation:

In a stoichiometry exercise we follow a series of steps.

Step 1: Write the balanced equation

2 CuCl₂ + 4 KI ⇄ 2 CuI + 4 KCl + I₂

Step 2: Look for the relationship between the data we have and the one we are looking for

According to the balanced equation, when 2 moles of CuCl₂ react 1 mole of I₂ is formed. This is the first relation. Secondly, we know that the molar mass of 253.80 g/mol, that is, every mole of I₂ has a mass of 253.80 g. This is the second relation.

Step 3: Apply relations to the data we have

0.5235mol(CuCl_{2}).\frac{1mol(I_{2})}{2mol(CuCl_{2})} .\frac{253.80g(I_{2})}{1mol(I_{2})} =66.43g(I_{2})

7 0
3 years ago
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