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Phoenix [80]
3 years ago
15

There is no sodium found in nature in pure form due to its reactivity. What does this say about activation energy of the reactio

n of sodium and water compared to the combustion of wax in a candle?
Chemistry
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

The activation energy of a chemical reaction is the energy that is required to be supplied for a chemical reaction to take place. The activation energy for the reaction of sodium is low compared to the energy released such that the reaction of sodium and water is spontaneous resulting in the melting of the sodium into liquid form

The activation energy of a candle wax is much higher, requiring the ignition of the wick which burns and in turn melts the candle wax to release vapors that burns alongside the wick to produce sooth carbon dioxide, carbon monoxide and water vapor and release of heat energy which also fuels further combustion of the candle wax and wick

Therefore, the activation energy of the candle wax and wick which require the heat of direct flame from an ignited matches is higher than the activation energy of sodium placed in a medium of water that reacts spontaneously without heat application

Explanation:

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How many grams of precipitate will be formed when 20.5 mL of 0.800 M
Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

<u>Step 1</u>: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

<u>Step 2:</u> Data given

Volume of 0.800 M  CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

<u>Step 3:</u> Calculate moles of CO(NO3)2

Moles CO(NO3)2  = Molarity * volume

Moles CO(NO3)2  = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step  6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0
3 years ago
In order to use the gas laws, all temperatures must be in the _ scale.
belka [17]

Kelvin. Kelvin is the scientific term used in temperature senses, and was created so everybody around the world would have the same measurements and not get confused about whether it was Fahrenheit or Celsius.

4 0
3 years ago
Read 2 more answers
Formula Silver(II) Oxide ​
Svetach [21]

Answer:

AgO

Explanation:

5 0
3 years ago
What are two things that you observed when you heated the mixture of tin and nitric acid over the Bunsen burner in the virtual l
Rus_ich [418]
1. The reaction for this would be:

Sn + 4 HNO₃ →  SnO₂ + 4 NO₂ + 2 H₂O

The first observation would be bubbling of the solution and brown acrid smoke is produced due to the presence of NO₂ gas. Another observation would be the presence of a white solid which is SnO₂.

2. Heating was required to get rid of the H₂O. When all moisture is gone, you weigh the sample. Afterwhich, you further heat it to get ride of the oxygen. By doing this, you would know the individual mass of each element. Then, you can solve for the empirical formula of the oxide of tin.
3 0
3 years ago
What substance is reduced in the reaction 2Fe 2+ + CI2 = 2Fe 3+ + 2CI- ?
atroni [7]

Answer:

$Cl_{2}$ is reduced in the reaction

Explanation:

The given reaction is

$2Fe^{2+}+Cl_{2} \to 2Fe^{3+}+2Cl^{-}$

The oxidation number of $Fe$ is changed from $+2\, \to \, +3$

$Fe^{2+} \to Fe^{3+}+e^{-}$

And The oxidation number of $Cl$ is changed from $0\, \to \, -1$

$Cl_{2}^{0} + 2e^{-} \to 2Cl^{-}$

Hence, $Fe^{2+}$ is oxidized and $Cl_{2}$ is reduced

5 0
3 years ago
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