Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL
Answer:
Chemists investigate the properties of matter at the level of atoms and molecules. They measure proportions and reaction rates in order to understand unfamiliar substances and how they behave, or to create new compounds for use in a variety of practical applications.
Explanation:
Answer:
The number of moles of xenon are 1.69 mol.
Explanation:
Given data:
Number of moles of xenon = ?
Volume of gas = 37.8 L
Temperature = 273 K
Pressure = 1 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will put the values in formula.
1 atm × 37.8 L = n × 0.0821 atm.L/ mol.K ×273 K
37.8 atm.L = n × 22.413 atm.L/ mol.
n = 37.8 atm.L / 22.413 atm.L/ mol.
n = 1.69 mol
The number of moles of xenon are 1.69.
The 2 represents that it is a double carbon bond
it looks like..
C-C = C-C
