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user100 [1]
3 years ago
7

What types of change if any can reach equilibrium

Chemistry
1 answer:
Lina20 [59]3 years ago
5 0
A chemical change. ....
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Determine the root-mean square speed of CO2 molecules that have an average kinetic energy of 4.2 x10-21 J per molecule.
strojnjashka [21]
<span>Determine the root-mean-square sped of CO2 molecules that have an average Kinetic Energy of 4.21x10^-21 J per molecule. Write your answer to 3 sig figs.
</span><span>
E = 1/2 m v^2 

If you substitute into this formula, you will get out the root-mean-square speed. 

If energy is Joules, the mass should be in kg, and the speed will be in m/s. 

1 mol of CO2 is 44.0 g, or 4.40 x 10^1 g or 4.40 x 10^-2 kg. 

If you divide this by Avagadro's constant, you will get the average mass of a CO2 molecule. 

4.40 x 10^-2 kg / 6.02 x 10^23 = 7.31 x 10^-26 kg 

So, if E = 1/2 mv^2 

</span>v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 115184.68 
Take the square root of that, and you get the answer 339 m/s.
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3 years ago
What is the theoretical yield of a reaction?
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When most liquids freeze, explain what happens to the motion and the space between the atoms.
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atoms relative motion slow down and begin to vibrate in place

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Which of the following material is the weakest thermal conducters
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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
2 years ago
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