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tatyana61 [14]
3 years ago
12

What molecules can cells break down for energy?

Chemistry
1 answer:
Orlov [11]3 years ago
6 0

Answer: I think the answer is C)

Explanation:

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true

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calculating the average atomic

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3 years ago
Given three (3.00) moles of gold (Au), how many grams do you have?
stellarik [79]
One mol of any element is the elements atomic mass. So you would look on the periodic table to find the atomic mass of gold. Which is 197g. Knowing that that is one mol of gold you would multiply 197 by three to get three mols of gold. Which is 591g. So the answer is c. I hope this helps!!
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3 years ago
What's a house hold item that has nuclear energy
lbvjy [14]

Answer:

Electricity, Weapons, Medicine · Food Treatments, ect.

3 0
3 years ago
Why is sugar considered an organic compound
Sliva [168]
Sugar is considered an organic compound,
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5 0
4 years ago
Determine the enthalpy change for the decomposition of calcium carbonate. CaCO3 (s) --> CaO (s) + CO2 (g) given the thermoche
Gnesinka [82]

Answer : The enthalpy change for the decomposition of calcium carbonate is, 178.1 kJ/mol

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 given main reaction is,

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)     \Delta H_1=65.02kJ/mol

(2) Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)    \Delta H_2=-113.8kJ/mol

(3) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_3=-393.5kJ/mol

(4) 2Ca(s)+O_2(g)\rightarrow 2CaO(s)    \Delta H_4=-1270.2kJ/mol

Now we are reversing reaction 1 and then adding reaction 1 and 2, we get :

(1) Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)     \Delta H_1=65.02kJ/mol

(2) CaCO_3(s)+H_2O(l)\rightarrow Ca(OH)_2(s)+CO_2(g)    \Delta H_2=113.8kJ/mol

The expression for enthalpy of change will be,

\Delta H=\Delta H_1+\Delta H_2

\Delta H=(65.02)+(113.08)

\Delta H=178.1kJ/mol

Thus, the enthalpy change for the decomposition of calcium carbonate is, 178.1 kJ/mol

7 0
4 years ago
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