The L- isomer serves as the enantiomer of the D- isomer, and given that the optical rotation of the D- isomer is + 13.5°, the L- isomer's optical rotation will have the same magnitude but an opposite sign, resulting in -13.5°.
Thus, the rotation of the racemic mixture will be equal to 0°.
- This occurs because a racemic mixture contains equal proportions of both enantiomers.
<span>The partial pressure of oxygen is 438.0 mmHg. The ideal gas equation is expressed as PV = nRT where P represents pressure, V denotes volume, n is the number of moles, R is the ideal gas constant (8.3144598 (L*kPa)/(K*mol)), and T signifies absolute temperature. To convert from Celsius to Kelvin, we have 43.4 + 273.15 = 316.55 K. For the pressure conversion from mmHg to kPa: 675.9 mmHg * 0.133322387415 = 90.11260165 kPa. When solving for n using the ideal gas equation, we derive n = PV / (RT) which provides n = 90.11260165 kPa * 16.2 L / (8.3144598 (L*kPa)/(K*mol) * 316.55 K)= 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol), resulting in n = 0.554656603 mol. Thus, we have 0.554656603 moles of gas particles. Next, we determine the contribution from oxygen. The atomic weight of oxygen is 15.999 g/mol, while argon is 39.948 g/mol, and the molar mass of O2 is 31.998 g/mol. We establish the relationships where M is the number of moles of O2, and 0.554656603 - M gives the number of moles of Ar. Setting up the equation: M * 31.998 + (0.554656603 - M) * 39.948 = 19.3, we solve for M resulting in 0.359424148 moles of oxygen out of 0.554656603 total moles. This leads to oxygen providing 0.359424148 / 0.554656603 = 0.648012024 or 64.8012024% of the total pressure of 675.9 mmHg. The partial pressure therefore calculates to 675.9 * 0.648012024 = 437.9913271 mmHg, rounded to 438.0 mmHg</span>
Answer:
9.69g
Explanation:
To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.
1 mole of gas takes up 22.4L at STP.
Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning
X moles of N2 = 7.744/22.4 = 0.346 moles
Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:
Molar Mass of N2 = 2x14 = 28g/mol
Number of moles N2 = 0.346 moles
Find the mass of N2 =?
Mass = number of moles × molar mass
Mass of N2 = 0.346 × 28
Mass of N2 = 9.69g
Hence, 7.744L of N2 consists of 9.69g of N2
Answer:
When HBCG and BCG^- are at the same concentration, the resulting color is green. This green shade initially becomes visible at a pH of 3.8.
Explanation:
HBCG serves as an indicator formed by dissolving solids in ethanol.
Since
Ka=[BCG−][H3O+][HBCG] When [BCG-] equals [HBCG], it follows that Ka = [H3O+].
<pWith a pH of 3.8,<pKa= [H3O+] = -antilog pH = -antilog (3.8)
Ka= 1.58 ×10^-4
Response:
9.9 ml of 0.200M NH₄OH(aq)
Reasoning:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?
1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution
1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)
=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters
