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mel-nik [20]
3 years ago
11

You are given two fatlike solid substances and determine that sample a has a higher melting point than sample

Chemistry
1 answer:
Alchen [17]3 years ago
5 0
Given that <span>sample a has a higher melting point than sample
b. Therefore, sample a is a longer chain of a </span><span>fatlike solid substance. It could also be that the bonds present in sample a is much stronger which will require more energy to break. Hope this answers the question.</span>
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Answer:

it is a weak acid

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3 0
2 years ago
a student wants to remove the salt from a mixture of sand and salt in order to get only person he adds water to the mixture why
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6 0
4 years ago
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Show your work. Write a balanced net ionic equation for the following reaction. H3PO4(aq) + Ca(OH)2(aq) Ca3(PO4)2(aq) + H2O(l)
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4 years ago
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How to draw Hess' Cycle for this question ?
NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0
3 years ago
Analytical Question: During Calvin cycle, photosynthesizing green algae are provided with CO2 containing radioactive oxygen (18O
harkovskaia [24]
....................
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