Find the mass of a metal cylinder that has a volume of 652 cm and a density of 21.45 g/cm
1 answer:
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Answer:
-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.
Explanation:
Mass of nitrogen triiodide = 20.0 g
Moles of nitrogen triiodide =
According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :
-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.
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Answer:
a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ, ΔSºrxn = 0.113 kJ/K
b) At 753.55 ºC or higher
c )ΔG = 1.8 x 10⁴ J
K = 8.2 x 10⁻²
Explanation:
a) C6H5−CH2CH3 ⇒ C6H5−CH=CH2 + H₂
ΔHf kJ/mol -12.5 103.8 0
ΔGºf kJ/K 119.7 202.5 0
Sº J/mol 255 238 130.6*
Note: This value was not given in our question, but is necessary and can be found in standard handbooks.
Using Hess law to calculate ΔHºrxn we have
ΔHºrxn = ΔHfº C6H5−CH=CH2 + ΔHfº H₂ - ΔHºfC6H5−CH2CH3
ΔHºrxn = 103.8 kJ + 0 kJ - (-12.5 kJ)
ΔHºrxn = 116.3 kJ
Similarly,
ΔGrxn = ΔGºf C6H5−CH=CH2 + ΔGºfH₂ - ΔGºfC6H5CH2CH3
ΔGºrxn= 202.5 kJ + 0 kJ - 119.7 kJ = 82.8 kJ
ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K
b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using
ΔGrxn = ΔHrxn -TΔS
we see that will happen when the term TΔS becomes greater than ΔHrxn since ΔS is positive , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that ΔºHrxn and ΔSºrxn remain constant at the higher temperature and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.
0 = 116 kJ -T (0.113 kJ/K)
T = 1026.5 K = (1026.55 - 273 ) ºC = 753.55 ºC
c) Again we will use
ΔGrxn = ΔHrxn -TΔS
to calculate ΔGrxn with the assumption that ΔHº and ΔSºremain constant.
ΔG = 116.3 kJ - (600+273 K) x 0.113 kJ/K = 116.3 kJ - 873 K x 0.113 kJ/K
ΔG = 116.3 kJ - 98.6 kJ = 17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )
Now for solving for K, the equation to use is
ΔG = -RTlnK and solve for K
- ΔG / RT = lnK ∴ K = exp (- ΔG / RT)
K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K x 873 K)) = 8.2 x 10⁻²
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Answer:
Option C. Energy Profile D
Explanation:
Data obtained from the question include:
Enthalpy change ΔH = 89.4 KJ/mol.
Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:
Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)
ΔH = Hp – Hr
Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.
If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.
Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.
Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.
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Instability
Explanation:
Isotopes decays because they are unstable. Stable isotopes do not decay.
- For every atomic nucleus, there is a specific neutron/proton ratio.
- This ratio ensure that a nuclide is stable.
- For example, fluorine F, is 10/9 stable.
- Any nucleus with a neutron/proton combination different from its stability ratio either too many neutrons or too many protons will become unstable.
- Such nuclide will split into one or more other nuclei with the emission of small particles of matter and considerable amount of energy.
Learn more:
Radioactive brainly.com/question/10125168
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