<h3>
Answer:</h3>
= 3.384 × 10^23 atoms
<h3>
Explanation:</h3>
We are given 18.016 g of sulfur
We are required to determine the number of atoms of sulfur
We will use the following steps;
<h3>Step 1: Determine the number of moles of sulfur </h3>
Number of moles = Mass ÷ molar mass
Molar mass of sulfur = 32.065 g/mol
Therefore;
Number of moles = 18.016 g÷ 32.065 g/mol
= 0.562 moles
<h3>Step 2: Determine the number of atoms </h3>
Using the Avogadro's constant;
1 mole of an element = 6.022 × 10^23 atoms
Therefore;
For 0.562 moles;
= 0.562 moles × 6.022 × 10^23 atoms
= 3.384 × 10^23 atoms
Therefore; there are 3.384 × 10^23 atoms in 18.016 g of Sulfur.
Answer:
0.8 mL of protein solution, 9.2 mL of water
Explanation:
The dilution equation can be used to relate the concentration C₁ and volume V₁ of the stock/undiluted solution to the concentration C₂ and volume V₂ of the diluted solution:
C₁V₁ = C₂V₂
We would like to calculate the value for V₁, the volume of the inital solution that we need to dilute to make the required solution.
V₁ = (C₂V₂) / C₁ = (2mg/mL x 10mL) / (25 mg/mL) = 0.8 mL
Thus, a volume of 0.8 mL of protein solution should be diluted with enough water to bring the total volume to 10 mL. The amount of water needed is:
(10 mL - 0.8 mL) = 9.2 mL
Answer:
- Elimination
- Elimination
- Zaitsev
- Zaitsev
- Carbocation
Explanation:
- The mechanism is generally accepted to always operate via an ELIMINATION step-wise process.
- The ELIMINATION mechanism process will always produce (after dehydration) a ZAITSEV style alkene as major product
- The driving force for the production of this ZAITSEV style alkene product is generally going to be determined by stability of the CARBOCATION
Elimination mechanism is the removal of two substituents from a molecule in either a one- or two-step mechanism
Carbocation is a molecule containing a positive charged carbon atom and three bonds
Molality is defined as the number of moles of solute in 1 kg of solvent.
In this case the solvent is water, therefore we need to find the number of moles of KBr in 1 kg of water.
Number of moles of KBr - 2.10 g / 119 g/mol = 0.0176 mol
mass of water - 897 g
number of moles of KBr in 897 g - 0.0176 mol
Therefore number of KBr moles in 1000 g - 0.0176 mol / 897 g x 1000 g = 0.0196 mol/kg
therefore molality of KBr is 0.0196 mol/kg