Gold is actually very soft. If rings were made of pure gold they would bend and become disfigured. Other elements are added to give the ring its stability.
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:
44.8 L
Explanation:
Ideal Gas Equation -
i.e.,
PV = nRT
where,
P = pressure
V = volume
n = moles
R = universal gas constant
T = temperature
Using the information given in the question, Volume of the gas can be calculated -
P = 101.3 kPa
V = ?
n = 2.00 moles
R = 8.31
T = 0 degree C = 273.15 K
Using the above data, and putting the data in the respective formula -
PV = nRT
101.3 kPa * V = 2.00 moles * 8.31 * 273.15 K
V = 44.8 L
Hence, the volume of the given gas = 44.8 L
