Living organisms use atoms from sugar molecules combined with other elements to form lipids and nucleic acids. Lipids are hydrophobic organic compounds composed of carbon and a high ratio of hydrogen to oxygen. Nucleic acids are an acidic polymer of nucleotides found or produced in the cell nucleus.
The answer should be 0.5 because 6/12 is 0.5. It gives u the key equation
Answer: 116 g of copper
Explanation:

where Q= quantity of electricity in coloumbs
I = current in amperes = 24.5A
t= time in seconds = 4.00 hr =
(1hr=3600s)

of electricity deposits 63.5 g of copper.
352800 C of electricity deposits =
of copper.
Thus 116 g of Cu(s) is electroplated by running 24.5A of current
Thus remaining in solution = (0.1-0.003)=0.097moles
<u>Answer:</u> The
for the reaction is -1835 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:

The intermediate balanced chemical reaction are:
(1)
( × 4)
(2)

The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B4%5Ctimes%20%28-%5CDelta%20H_1%29%5D%2B%5B1%5Ctimes%20%5CDelta%20H_2%5D)
Putting values in above equation, we get:

Hence, the
for the reaction is -1835 kJ.
<u>61.25 grams</u> of CO can be formed from 35 grams of oxygen.
The molecular mass of oxygen is <u>16 gmol⁻¹</u>
The molecular mass of carbon monoxide is<u> 28 gmol⁻¹</u>
Explanation:
The molar mass of carbon monoxide is molar mass of C added to that of O;
12 + 16 = 28
= 28g/mol
The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol
Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;
35/32 * 2
= 70/32 moles
Then multiply by the molar mass of carbon monoxide;
70/32 * 28
= 61.25 g