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natima [27]
2 years ago
15

The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C u

pper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed. is halved. has its sign changed. is unchanged.
Chemistry
1 answer:
GenaCL600 [577]2 years ago
7 0

Answer:

the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

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All amino acids that are found in proteins, except for proline, contain a(n); Group of answer choices amino group carbonyl group
Jobisdone [24]

Answer:

amino group

Explanation:

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Proline, which is the only cyclic amino acid, is also the only amino acid that forms a secondary amine group i.e. loss of hydrogen atoms in its amine group when in a protein structure. This means that when in a protein, PROLINE does not have an AMINE GROUP.

7 0
2 years ago
A 0.300 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 66,
Alja [10]

Answer:

X=600 degrees celsius

Explanation:

We can evaluate by using the equation ==> Q=m(c)(delta T)

-We know the mass is 0.3 g

-Q= 66,300 J

-C= 390 j/g 0C

-Therefore, we need to solve for delta T (which will be represented by x)

 Q= mct

66,300 J= 0.3 g (390 J/g 0C) (t)

The unit of measurement (g-grams) can be crossed out on the left side of our expression.

66,300 J=0.3(390 J 0C) (t)

-If we convert into a proportional relationship:

66,300 J                117 J/0C x

------------       =        -----------

117 J/0C                 117 J/0C

-Joules on both sides can be crossed off

*117 J/0C came from the product in terms of 0.3(390 J 0C)

*If we find the quotient of 66,300 J and 117 J/0C, we get 566.6 repeating

-You may keep the answer as it is, although for simplicity reasons, we can round to the nearest hundred.

-Therefore, x=600.

-----Remember, x represented the delta t in our equation. (I made notice of that when evaluating).

Hope this helps! :)

3 0
1 year ago
identify A and B, isomers of molecular formula C3H4Cl2, from the given 1H NMR data: Compound A exhibits peaks at 1.75 (doublet,
siniylev [52]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

3 0
2 years ago
A 92.8 gram sample of CuSO4•5H2O.(copper sulfate pentahydrate) is heated until water is released. How many grams of water were r
Ivan

Mass of water released =

92.8 g CuSO_{4}.5H_{2}O×\frac{5 * 18 g H_{2}O}{249.68 g CuSO_{4}.5H_{2}O}

= 33.45 g H_{2}O

7 0
2 years ago
Read 2 more answers
Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a
JulijaS [17]

Answer : The correct option is, 30.9^oC

Explanation :

Formula used :

q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

q = heat released = 24 KJ

m = mass of bomb calorimeter = 1.30 Kg

c = specific heat = 3.41J/g^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 25.5^oC

Now put all the given values in the above formula, we get  the final temperature of the calorimeter.

q=m\times c\times (T_{final}-T_{initial})

24KJ=1.30Kg\times 3.41J/g^oC\times (T_{final}-25.5)^oC

T_{final}=30.9^oC

Therefore, the final temperature of the calorimeter is, 30.9^oC

5 0
2 years ago
Read 2 more answers
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