Answer:
<h3>A. Epimers </h3>
Explanation:
<h2>Hope it help ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️</h2>
Answer:
Volume of sample after droping into the ocean=0.0234L
Explanation:
As given in the question that gas is idealso we can use ideal gas equation to solve this;
Assuming that temperature is constant;
Lets
and
are the initial gas parameter before dropping into the ocean
and
and
are the final gas parameter after dropping into the ocean
according to boyle 's law pressure is inversly proportional to the volume at constant temperature.
hence,
![P_1V_1=P_2V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2)
P1=1 atm
V1=1.87L
P2=80atm
V2=?
After putting all values we get;
V2=0.0234L
Volume of sample after droping into the ocean=0.0234L
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.
![1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L](https://tex.z-dn.net/?f=1%20mL%20%3D%200.001%20L%5C%5C31.6%20mL%20%3D%2031.6%20mL%20%5Ctimes%20%5Cfrac%7B0.001%20L%7D%7B1%20mL%7D%5C%5C%3D%200.0316%20L)
Temperature = ![91^{o}C = (91 + 273) K = 364 K](https://tex.z-dn.net/?f=91%5E%7Bo%7DC%20%3D%20%2891%20%2B%20273%29%20K%20%3D%20364%20K)
As molar mass of
is 32 g/mol. Hence, the number of moles of
are calculated as follows.
![No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol](https://tex.z-dn.net/?f=No.%20of%20moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C%3D%20%5Cfrac%7B0.023%20g%7D%7B32%20g%2Fmol%7D%5C%5C%3D%200.00072%20mol)
Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.
![PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm](https://tex.z-dn.net/?f=PV%20%3D%20nRT%5C%5CP%20%20%5Ctimes%200.0316%20L%20%3D%200.00072%20mol%20%5Ctimes%200.0821%20L%20atm%2Fmol%20K%20%5Ctimes%20364%20K%5C%5CP%20%3D%20%5Cfrac%7B0.00072%20mol%20%5Ctimes%200.0821%20L%20atm%2Fmol%20K%20%5Ctimes%20364%20K%7D%7B0.0316%20L%7D%5C%5C%3D%200.681%20atm)
Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.