Thus, it follows that after 4 to 5 half-lives, the plasma concentrations of a given drug will be below a clinically relevant concentration and thus will be considered eliminated. Conversely, the accumulation of a drug can reach a steady-state during an infusion
Explanation:
Methane molecule is depicted here
Answer:
1.47 mol/L
Explanation:
Molarity is given as,
Molarity = Moles / Vol in L ------- (1)
Moles of CaCl₂,
Moles = Mass / M.Mass
Moles = 535 g / 110.98 g/mol
Moles = 4.82 mol
Now, putting values in eq. 1.
Molarity = 4.82 mol / 3.28 L
Molarity = 1.47 mol/L
Answer:
The final pressure in the container at 0°C is 2.49 atm
Explanation:
We apply the Ideal Gases law to know the global pressure.
We need to know, the moles of each:
P He . V He = moles of He . R . 273K
(1atm . 4L) / R . 273K = moles of He → 0.178 moles
P N₂ . V N₂ = moles of N₂ . R . 273K
(1atm . 6L) / R . 273K = moles of N₂ → 0.268 moles
P Ar . V Ar = moles of Ar . R . 273K
(1atm . 10L) / R . 273K = moles of Ar → 0.446 moles
Total moles: 0.892 moles
P . 8L = 0.892 mol . R . 273K
P = ( 0.892 . R . 273K) / 8L = 2.49 atm
R = 0.082 L.atm/mol.K
Answer:
0.0451 mol FeCl₂
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
5.72 g FeCl₂
<u>Step 2: Identify Conversions</u>
Molar Mass of Fe - 55.85 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of FeCl₂ - 55.85 + 2(35.45) = 126.75 g/mol
<u>Step 3: Convert</u>
<u />
= 0.045128 mol FeCl₂
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.045128 mol FeCl₂ ≈ 0.0451 mol FeCl₂
