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gayaneshka [121]
4 years ago
11

Given:

Chemistry
1 answer:
Ivanshal [37]4 years ago
6 0
It depends on what kind of reaction it is. it could be 10 moles too, but it can be less, for example, if the products looked like this
CO2+2 H2O the answer would be:

x mol CO2 - 10 mol H20
1 mol CO2 - 2 mol H2O

x =  \frac{1 \times 10}{2}  = 5 \: mol
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4 years ago
What did Rutherford's model of the atom look like? a dense nucleus surrounded by electrons a random mixture of electrons and pro
RSB [31]

Answer:

Explanation:  Rutherford model, also called Rutherford atomic model, nuclear atom, or planetary model of the atom, description of the structure of atoms proposed (1911) by the New Zealand-born physicist Ernest Rutherford. The model described the atom as a tiny, dense, positively charged core called a nucleus, in which nearly all the mass is concentrated, around which the light, negative constituents, called electrons, circulate at some distance, much like planets revolving around the Sun. Hope that helps!

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1. How are atoms held together? What is the attraction between?
barxatty [35]

Answer:

the atom is held together by the electrostatic attraction between the positively charged nucleus and the negatively charged electrons surrounding it, the stability within chemical bonds is also due to electrostatic attractions.

Explanation:

3 0
3 years ago
Read 2 more answers
Power plants can be powered by fossil fuels, such as coal or natural gas, or nuclear power to produce steam to turn turbines to
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7 0
3 years ago
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Consider the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) Using standard thermodynamic data at 298K, calculate the entropy chang
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.

5 0
3 years ago
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