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GrogVix [38]
2 years ago
12

What is the resulting pressure when you transfer gas from cylinder with a temperature of 22°C and the pressure of four atm into

a container with a temperature of 65°C? Assume volume remains constant.
Chemistry
1 answer:
Hatshy [7]2 years ago
4 0
Answer:
             P₂  =  4.58 atm

Solution:
               According to <span>Gay-Lussac's law ," the pressure of a given mass of an ideal gas at constant volume is directly proportional to absolute temperature".
For initial and final states of a gas, the relation is given as,

                                           P</span>₁ / T₁  =  P₂ / T₂   ----- (1)

Data Given;
                   P₁  =  4 atm

                   T₁  =  22 °C + 273  =  295 K

                   P₂  =  ??

                   T₂  =  65 °C + 273  =  338 K

Now, solving equation 1 for P₂,

                  P₂  =  P₁ T₂ / T₁

Putting values,

                  P₂  =  (4 atm × 338 K) ÷ 295 K

                  P₂  =  4.58 atm
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Answer:

No reaction

Explanation:

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However, zinc can replace copper from its salt as:

Zn_{(s)}+Cu(NO_3)_2_{(aq)}\rightarrow Cu_{(s)}+Zn(NO_3)_2_{(aq)}

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Refrigerators are usually kept at about 5°C, while room temperature is about 20°C. if you were to take an empty sealed 2 liter s
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c) No, because Celsius is not an absolute temperature scale

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5 0
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Read 2 more answers
Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---&gt; CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu
Nitella [24]

ΔH = +438 kJ  

We have three equations:  

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(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ  

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ  

From these, we must devise the target equation:  

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?  

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The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.  

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.  

(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +92 kJ  

Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

<em>Reverse Equation (III).</em>  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ  

Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.  

You need ³/₂O₂ on the left.  

Multiply <em>Equation (II) by three</em>.  

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ  

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.  

When you add equations, you add their Δ<em>H</em> values.  

_______________________________________

We get the target equation (IV):  

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂;                                    ΔH = +  92 kJ  

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ  

(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O;                             ΔH =   -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  


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