Answer:
2Au₂S₃ + 6H₂ → 4Au + 6H₂S
Explanation:
Balancing:
2Au₂S₃ + 6H₂ → 4Au + 6H₂S
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
The answer to problem is [He] 2s1. Hope it helps
Answer:
V2 = 3.11 x 105 liters
Explanation:
Initial Volume, V1 = 2.16 x 105 liters
Initial Temperature, T1 = 295 K
Final Temperature, T2 = 425 K
Final Volume, V2 = ?
These quantities are related by charle's law and the equation of the law is given as;
V1 / T1 = V2 / T2
V2 = T2 * V1 / T1
V2 = 425 * 2.16 x 105 / 295
V2 = 3.11 x 105 liters
Answer:
See attached picture.
Explanation:
Hello,
In this case, for the given name, you can verify the structure on the attached picture, wherein you can see verify the presence of both the ethyl and methyl radicals at the third carbon as well as the triple bond at the first carbon.
Best regards.
