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antiseptic1488 [7]
3 years ago
15

Which reaction is represented by the net ionic equation for the combination of aqueous solutions of LiNO3 and NaC2H3O2?

Chemistry
1 answer:
dangina [55]3 years ago
7 0

Answer:

This question is incomplete

Explanation:

This question is incomplete but the completed question is in the attachment below. To determine the net ionic equation, one must first write the complete equation which is shown below

LiNO₃ + NaC₂H₃O₂ ⇒  LiC₂H₃O₂ + NaNO₃

The ionic equation represented from the above equation will be

Li¹⁺ NO₃⁻ + Na¹⁺ C₂H₃O₂¹⁻  ⇒  Li¹⁺ C₂H₃O₂¹⁻ + Na¹⁺NO₃¹⁻

Because the resultant NaNO₃ formed is an ionic compound (which does not form a molecule), the correct option from the options provided is the first option

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Joseph divides 8.64 by 2.0. How many significant figures should his answer have?
nadezda [96]

Answer:

2 significant figures

Good luck with your work! :)

6 0
3 years ago
How many grams of potassium carbonate are needed to make 300ml of a 4.5M solution?
kenny6666 [7]

Answer:

186.3g

Explanation:

4.5moles of K₂CO₃  is in 1000ml

? moles of K₂CO₃ is in 300 ml

(4.5 × 300)/ 1000 = 1.35 moles of K₂CO₃

1 mole of K₂CO₃ = (39 × 2) + 12 + (16 × 3) = 78 + 12 + 48 = 138g

1.35 moles of K₂CO₃  = ?

= (1.35 × 138)/1 = 186.3g

4 0
3 years ago
Pls hurry
eimsori [14]

Answer:

plusses give me brainliest after this

The answer is: 1. periodic table

4 0
3 years ago
A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago was the artif
noname [10]

Answer:

The age of the sample is 4224 years.

Explanation:

Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = [A_o]

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = [A]=0.06[A_o]

Half life of the carbon-14 = t_{1/2}=5730 years

k=\frac{0.693}{t_{1/2}}

[A]=[A_o]\times e^{-kt}

[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}

0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}

Solving for t:

t = 4223.71 years ≈ 4224 years

The age of the sample is 4224 years.

8 0
4 years ago
(8.6 1029) 7.4 X1029)
andrew11 [14]

Answer:

65563.914234

Explanation:

8.61029 x 7.4 x 1029

63.716146 x 1029

multiply

= 65563.914234

3 0
3 years ago
Read 2 more answers
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