The patient needs to receive 285.71 ml.
1000 ml contains 70 gr of glucose.
x contains 20 gr of glucose.
x=1000*20/70
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.
Answer:
The molality is 1.15 m.
Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.
Calculate moles of H₂SO₄ from molarity:
C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles
Mass of solvent (water) based on density:
m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg
Therefore, molality is:
m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m
In a water molecule, the sharing of electrons occurs between the oxygen and hydrogen atoms within covalent bonds; however, this sharing is unequal. The oxygen atom holds a stronger pull on the electrons compared to the hydrogen atoms in the bond.
Response:
Here's my calculation
Clarification:
Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We need to determine the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's compile all the information in one location.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial values
The graph below visualizes the initial concentrations as plotted on the vertical axis.