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Sveta_85 [38]
2 years ago
10

Consider the formation of the three solutionsshown in the table.

Chemistry
1 answer:
bazaltina [42]2 years ago
8 0

Answer:

sorry please can you snap the diagram ...the question is not clear to my understanding

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Light moves in a straight line except at surfaces between different transparent materials, where its path bends.
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3 years ago
The solubility table helps predict the products of what kind of reactions?
Ganezh [65]

Answer:

B. double replacement

Explanation:

Rate, or give brainliest if you can :)

3 0
2 years ago
If a 6 ohm wire is connected to a 10 volt battery, what will the current be?
lidiya [134]

Answer:

60 amperes

Explanation:

5 0
2 years ago
Read 2 more answers
A student decomposed 3.67g of copper (ii) hydroxide into copper (ii) oxide. how many ml of 3m h2so4 is need to react with all th
vredina [299]
In the presence of heat, copper (II) hydroxide decomposes in to copper (II) oxide. 
Cu(OH)₂ (s) ----> CuO (s) + H₂O (l)
upon decomposition, water is removed from Cu(OH)₂
the amount of Cu(OH)₂ decomposed - 3.67 g
number of moles of Cu(OH)₂  - 3.67 g / 97.5 g/mol = 0.038 mol
stoichiometry of Cu(OH)₂ to CuO is 1:1
therefore number of CuO moles formed are - 0.038 mol
CuO reacts with sulfuric acid to form CuSO₄ 
CuO + H₂SO₄ ---> CuSO₄ + H₂O
stoichiometry of CuO to H₂SO₄ is 1:1
therefore number of H₂SO₄ moles that should react is 0.038 mol
the molarity of H₂SO₄ is 3M
this means that in 1000 ml - 3 mol of H₂SO₄ present 
so if 3 mol are present in 1000 ml 
then volume for 0.038 mol = 1000/3 * 0.038 
                                           = 12.67 ml
6 0
3 years ago
What is oxidized in a galvanic cell with aluminum and gold electrodes ?
k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V

Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Notice that the overall cell potential upon summing is:

E_{cell}=1.66 V + 1.50 V=3.16 V

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0
3 years ago
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