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musickatia [10]
3 years ago
5

PLSSS HELP MEE 100 POINTS!!!!!!! How many atoms of each element does magnesium hydroxide, Mg(OH)2, contain? Magnesium = 1 atom,

Oxygen = 2 atoms, Hydrogen = 2 atoms Magnesium = 2 atom, Oxygen = 2 atoms, Hydrogen = 2 atoms Magnesium = 1 atom, Oxygen = 1 atoms, Hydrogen = 2 atoms Magnesium = 2 atom, Oxygen = 1 atoms, Hydrogen = 2 atoms
Chemistry
2 answers:
Basile [38]3 years ago
5 0

It would be Magnesium = 1 atom, Oxygen = 2 atoms, Hydrogen = 2 atoms. It has 5 atoms total. There is 1 magnesium and you multiply each element by the the 2 outside the parenthesis to get 2 oxygen and 2 hydrogen.

kompoz [17]3 years ago
3 0

how many atoms Mg(OH)2 contains? choices are:

Magnesium = 1 atom, Oxygen = 2 atoms, Hydrogen = 2 atoms

Magnesium = 2 atom, Oxygen = 2 atoms, Hydrogen = 2 atoms

Magnesium = 1 atom, Oxygen = 1 atoms, Hydrogen = 2 atoms

Magnesium = 2 atom, Oxygen = 1 atoms, Hydrogen = 2 atoms


look at Mg(OH)2, there should be 1x Mg, 2x O n 2x H

so ans is A.  Magnesium = 1 atom, Oxygen = 2 atoms, Hydrogen = 2 atoms


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balandron [24]

Answer:

the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.

Explanation:

3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)

∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K

∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.

For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,

then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.

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c) 9.03 x 10^23

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Then use that, to find the number of moles in Aluminum.

Then use Avogadro's number which is 6.02 * 10^23

After that, write all of that down with dimensional analysis.

40.5 g * 1 mol/ 27.0 g of Al  * 6.02 x 10^23 / 1mol

As your final answer, you will get 9.03 * 10^23 atoms with sig figs.

Hope it helped!

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