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MA_775_DIABLO [31]
3 years ago
12

5.In the atom, the electrons can be found in the

Chemistry
1 answer:
HACTEHA [7]3 years ago
5 0

Answer:

Probably D, Electron Cloud.

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Predict at least two circumstances that would cause a system in equilibrium to change the concentrations of its reactants and pr
Masja [62]
At the point when a response achieves a condition of concoction balance under a specific arrangement of conditions, no further changes in the groupings of reactants and items happen. On the off chance that a change is made in the conditions under which the framework is at harmony, substance change will happen so as to build up another balance. The components that can impact harmony are an adjustment in focus, change in weight (or volume), and change in temperature.
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How many hydrogen atoms are in 8.80 mol of ammonium sulfide
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Assuming that the ammonium sulfide formula is (NH4)2S then you can see that there are 2 nitrogen, 8 hydrogen and 2 sulfur atoms for every ammonium sulfide. If the amount of ammonium sulfide is 8.9 moles, then the number of hydrogen atoms should be: 8/1 * 8.9 mol= 71.2 moles
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Is the formation of sulfur tetrafluoride monoxide endothermic or exothermic?
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3 years ago
The molal boiling-point and freezing-point constants are different for different solvents.
RoseWind [281]

Answer:

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Explanation:

8 0
3 years ago
Read 2 more answers
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

3 0
3 years ago
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