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Whitepunk [10]
3 years ago
11

I really need some help pleasee

Chemistry
1 answer:
vlabodo [156]3 years ago
6 0

Answer:

=<em>2.1 moles of carbon</em>

Explanation:

From avogadro's number,

<em>6</em><em>.</em><em>0</em><em>2</em><em>×</em><em>1</em><em>0</em><em>^</em><em>2</em><em>3</em><em> </em><em>a</em><em>t</em><em>o</em><em>m</em><em>s</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>c</em><em>o</em><em>n</em><em>t</em><em>a</em><em>i</em><em>n</em><em>e</em><em>d</em><em> </em><em>i</em><em>n</em><em> </em><em>1</em><em> </em><em>m</em><em>o</em><em>l</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>C</em>

<em>1</em><em>.</em><em>2</em><em>5</em><em>×</em><em>1</em><em>0</em><em>^</em><em>2</em><em>4</em><em> </em><em>a</em><em>t</em><em>o</em><em>m</em><em>s</em><em> </em><em>w</em><em>i</em><em>l</em><em>l</em><em> </em><em>b</em><em>e</em><em> </em><em>c</em><em>o</em><em>n</em><em>t</em><em>a</em><em>i</em><em>n</em><em>e</em><em>d</em><em> </em><em>i</em><em>n</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>(</em><em>1.25×10^24</em><em>)</em><em>÷</em><em>(</em><em>6.02×10^23</em><em>)</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>2</em><em>.</em><em>1</em><em> </em><em>moles</em><em> </em><em>of</em><em> </em><em>carbon</em>

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1.302  moles of carbon dioxide would have to be added

<h3>Further explanation</h3>

The equilibrium constant is the value of the product in the equilibrium state of the substance in the right (product) divided by the substance in the left (reactant) with the exponents of each reaction coefficient

The equilibrium constant is based on the concentration (Kc) in a reaction

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\displaystyle Kc=\frac{0.75.0.75}{0.2.0.2}\\\\Kc=14.1

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Reaction :

                CO +H₂O ⇔ CO₂ + H₂

initially      0.2   0.2       0.75+x  0.75

reaction    0.1    0.1        0.1         0.1

product     0.3   0.3       0.65+x    0.65

\displaystyle Kc=\frac{(0.650+x)(0.65)}{0.3.0.3}\\\\14.1(0.09)=0.4225+0.65x\\\\1.269-0.4225=0.65x\\\\0.8465=0.65x\\\\x=1.302

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