Answer:
<u>When small organic molecules bind together, they form larger molecules called biological macromolecules.</u>Biological macromolecules are important cellular components and perform a wide array of functions necessary for the survival and growth of living organisms. The four major classes of biological macromolecules are carbohydrates, lipids, proteins, and nucleic acids.
(i hope this helps)
Answer : The partial pressure of
and
are, 84 torr and 778 torr respectively.
Explanation : Given,
Mass of
= 15.0 g
Mass of
= 22.6 g
Molar mass of
= 197.4 g/mole
Molar mass of
= 32 g/mole
First we have to calculate the moles of
and
.
![\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DC_2HBrClF_3%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DC_2HBrClF_3%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DC_2HBrClF_3%7D%3D%5Cfrac%7B15.0g%7D%7B197.4g%2Fmole%7D%3D0.0759mole)
and,
![\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DO_2%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DO_2%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DO_2%7D%3D%5Cfrac%7B22.6g%7D%7B32g%2Fmole%7D%3D0.706mole)
Now we have to calculate the mole fraction of
and
.
![\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DC_2HBrClF_3%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DC_2HBrClF_3%7D%7B%5Ctext%7BMoles%20of%20%7DC_2HBrClF_3%2B%5Ctext%7BMoles%20of%20%7DO_2%7D%3D%5Cfrac%7B0.0759%7D%7B0.0759%2B0.706%7D%3D0.0971)
and,
![\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DO_2%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DO_2%7D%7B%5Ctext%7BMoles%20of%20%7DC_2HBrClF_3%2B%5Ctext%7BMoles%20of%20%7DO_2%7D%3D%5Cfrac%7B0.706%7D%7B0.0759%2B0.706%7D%3D0.903)
Now we have to partial pressure of
and
.
According to the Raoult's law,
![p^o=X\times p_T](https://tex.z-dn.net/?f=p%5Eo%3DX%5Ctimes%20p_T)
where,
= partial pressure of gas
= total pressure of gas
= mole fraction of gas
![p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T](https://tex.z-dn.net/?f=p_%7BC_2HBrClF_3%7D%3DX_%7BC_2HBrClF_3%7D%5Ctimes%20p_T)
![p_{C_2HBrClF_3}=0.0971\times 862torr=84torr](https://tex.z-dn.net/?f=p_%7BC_2HBrClF_3%7D%3D0.0971%5Ctimes%20862torr%3D84torr)
and,
![p_{O_2}=X_{O_2}\times p_T](https://tex.z-dn.net/?f=p_%7BO_2%7D%3DX_%7BO_2%7D%5Ctimes%20p_T)
![p_{O_2}=0.903\times 862torr=778torr](https://tex.z-dn.net/?f=p_%7BO_2%7D%3D0.903%5Ctimes%20862torr%3D778torr)
Therefore, the partial pressure of
and
are, 84 torr and 778 torr respectively.
Answer:
Chemical energy is converted into thermal and mechanical energy while kinetic energy into potential energy.
Explanation:
As the breakdown of food started, the chemical energy present in the food substances converted into thermal energy and mechanical energy. Thermal or heat energy is used to maintain the body's temperature while mechanical energy is used in other activities such as movement of muscles. Kinetic energy is also converted into potential energy when the object moves from top of a place such as hill.
<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
Answer:
The correct answer is 1.21 L.
Explanation:
Based on the given information, the reaction will be,
CS2 (l) + 3Cl2 (g) ⇒ CCl4 (l) + S2Cl2 (l)
By using the standard values of the substances, the standard enthalpy of the reaction is,
ΔH° = [(-139.5) + (-58.5) – 0 – (87.3)] kJ/mol
= -285.3 kJ/mol
The amount of heat evolved for 3 moles of chlorine reacted us 285.3 kJ.
Now the number of moles of chlorine needed to react to produce 5.00 kJ is,
= 5.00 kJ × 3 mol Cl2/285.3 kJ
= 0.0526 mol Cl2
Now the volume of chlorine gas at 27degree C and 812 mmHg will be,
Volume = 0.0526 mol Cl2 × 0.0821 Latm/mol K × 300 K/ 1.07 atm
= 1.21 L