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3 years ago

If you designed your own experiment to prove the law of conservation of mass, what conditions would be required?

1 answer:

8 0

1) No. of atoms of all elements participating in the reaction are to be balanced.
2) Pressure Change should be noted (because, if ∆P≈0, it shows the reaction is in equilibrium).
3) The theoretical representation of the reaction should be balanced well.

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What is the net ionic equation for 2AgNO3 + 2NaOH = Ag2O + 2NaNO3 + H2O

soldi70 [24.7K]

Answer:

2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

Explanation:

Step 1: RxN

2AgNO₃ + 2NaOH → Ag₂O + 2NaNO₃ + H₂O

Step 2: Define states of matter

2AgNO₃ (aq) + 2NaOH (aq) → Ag₂O (s) + 2NaNO₃ (aq) + H₂O (l)

Step 3: Total Ionic Equation

2Ag⁺ (aq) + 2NO₃⁻ (aq) + 2Na⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + 2Na⁺ (aq) + 2NO₃⁻ (aq) + H₂O (l)

Step 4: Cancel out spectator ions

2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

7 0

3 years ago

What affects the density of a substance? Select all that apply. Mass Gravity Volume Weight

Vlada [557]

Mass and Volume!

Hope this helps. :)

7 0

3 years ago

"An acid is a substance that forms hydrogen ions when dissolved in water." this is an example of-

ValentinkaMS [17]

Hmm. This is very unspecific but it could either be indicating the pH of the solution or just saying it is an aqeous solution.

5 0

3 years ago

Read 2 more answers

Why should a new splint be used for each compound?

Inessa [10]

a cord could support it.

7 0

3 years ago

Read 2 more answers

If 16.00 g of O₂ reacts with 80.00 g NO, what is the excess reactant? * A.O₂ B.NO

Fed [463]

Answer:

B. NO

Explanation:

In each mole of NO, the weight of the oxygen is 16.00 g; the weight of the nitrogen is 14.01 g. Then in 80 g of NO, we have 42.65 g of oxygen and 37.35 g of nitrogen.

In each mole of NO₂, the weight of the oxygen is 2(16.00) = 32 g. The weight of the nitrogen is 1(14.01) = 14.01 g.

From the available oxygen, we can produce ...

(42.65 g +16 g)/(32.00 g/mol) = 1.83 mol of NO₂

From the available nitrogen, we can produce ...

(37.35 g)/(14.01 g/mol) = 2.67 mol of NO₂

Clearly, the reaction is limited by the amount of available oxygen.

NO is the excess reactant.