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ruslelena [56]
3 years ago
9

put the following in order from LARGEST TO SMALLEST nuclueus,molecule,electron,atom,proton and a teacher

Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
3 0

Answer:

electron, proton, nucleus, atom, molecule teacher

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Given the following equation: 2 KClO 3 --> 2 KCl + 3 O 2
SCORPION-xisa [38]

Explanation:

12.0 mol KClO3 × (3 mol O2/2 mol KClO3)

= 18.0 mol O2

6 0
3 years ago
Please explain:) I would really appreciate a step by step explanation if possible.
UNO [17]

The enthalpy change of the reaction, ΔH = -311 kJ

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

<h3>What is the enthalpy change for the reduction of ethyne to form ethane?</h3>

The enthalpy change for the reaction is obtained from the summation of the enthalpies of the reactions of the intermediate steps according to Hess's law.

The equation of the reaction is given below:

  • C₂H₂ + 2 H₂ → C₂H₆

The enthalpy of the reaction, ΔH = ΔH₁ + 2ΔH₂ + (-ΔH₃)

ΔH = {(-1299) + (2 * -286) + (1560)}Kj

ΔH = -311 kJ

The equation for the methanation reaction is given below:

3 H₂O + CO → CH₄ + H₂O

The enthalpy for the methanation reaction is as follows:

ΔH = 1.5ΔH₁ + 0.5*(-ΔH₂) + ΔH₃ + -ΔH₄

ΔH = (-483.6 * 1.5) + (0.5 * 221.0) + (-802.7) + (393.5)

ΔH = -1024.1 kJ/mol

Molar mass of CO = 28 /mol

Enthalpy change involved in the reaction of 300 g of CO = 300/28 * -1024.1 kJ/mol

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

In conclusion, the enthalpy changes are calculated from the enthalpy values of the  intermediate reactions.

Learn more about enthalpy changes at: brainly.com/question/26991394

#SPJ1

7 0
2 years ago
A substance which melts/boils at a fixed temperature is a pure substance.<br><br> True<br> False
ololo11 [35]
True

when a substance is impure, it boils over a range of temperature rather than a specific temperature
3 0
3 years ago
Read 2 more answers
What causes the energy transformation that occurs when the brake pads on a bicycle
makkiz [27]
Your answer would be D friction.
5 0
3 years ago
Consider the following intermediate chemical equations. When you form the final chemical equation, what should you do with CO? C
-Dominant- [34]

Answer:

Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.

Explanation:

The CO appears twice hence in he intermediate reaction it only forms path of the enabling reagents and it further reacts to form the final product. Accounting for the CO in the intermediate reaction that undergoes further reaction will impact on the stoichiometry of the reaction.

5 0
3 years ago
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