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Elis [28]
3 years ago
9

What is happening to the temperature of the substance BEFORE AND AFTER the phase changes?

Chemistry
1 answer:
Neporo4naja [7]3 years ago
5 0
First off, you must realize that the phase changes are marked by the points B and D on the graph. They are level because all of the energy (or heat) being added is being consumed by the physical process. So The temperature is increasing before the phase change, and after the phase change. The moments before and after are represented by points A, C, and E.
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Sulfer is a nonmetal
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2 years ago
Enter the net ionic equation for this reaction. Express your answer as a net ionic equation. Identify all of the phases in your
Lisa [10]

Answer:

2H+(aq) + 2OH-(aq) → 2H2O(l)

Explanation:

Step 1: The balanced equation

2HCl(aq)+Ca(OH)2(aq) → 2H2O(l)+CaCl2(aq)

This equation is balanced, we do not have the change any coefficients.

Step 2: The netionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.

2H+(aq) + 2Cl-(aq) + Ca^2+(aq) + 2OH-(aq) → 2H2O(l) + Ca^2+(aq) + 2Cl-(aq)

After canceling those spectator ions in both side, look like this:

2H+(aq) + 2OH-(aq) → 2H2O(l)

6 0
3 years ago
Make a prediction about the relationship between electrons and molecular shapes ​
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Answer:

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Explanation:

Hope you have a great day :)

6 0
2 years ago
If you were converting from milli- to centi- units, would you move the decimal point to the left or the right
yan [13]

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6 0
3 years ago
For the reaction CO+2H2=CH3OH at 700 K, equilibrium concentrations are [H2]=0.072 M, [CO]= 0.020M, and [CH3OH]= 0.030 M. Calcula
bekas [8.4K]

The balanced equation for the reaction is

CO(g) + 2H₂(g) ⇄ CH₃OH(g)

 

The given concentrations are at equilibrium state. Hence we can use them directly in calculation with the expression for the equilibrium constant, k. expression for k can be written as

   k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²

</span>[H₂<span>]=0.072 M
[CO]= 0.020M
[CH</span>₃OH]= 0.030 M

 

From substitution,

   k = 0.030 M / 0.020 M x (0.072 M)²

   k = 289.35 M⁻²

<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².

<span> </span>

3 0
3 years ago
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