Answer: 70.0°C
Explanation:
Quantity of heat = Mass * Specific heat * Change in temperature
Quantity of heat = 104.6 KJ
Mass = 500.0 g
Specific heat of water is 4.18 J/g°C
Change in temperature assuming final temperature is x = x - 20
Units should be in grams and joules:
104,600 = 500 * 4.18 * (x - 20)
104,600 = 2,090 * (x - 20)
x - 20 = 104,600/2,090
x = 104,600/2,090 + 20
x = 69.8
= 70.0°C
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:
697 g
Explanation:
Ethanol (C₂H₅OH) and butanoic acid (C₃H₇COOH) react to form ethyl butanoate (C₃H₇COOC₂H₅) and water (H₂O).
C₂H₅OH + C₃H₇COOH → C₃H₇COOC₂H₅ + H₂O
The molar ratio of C₂H₅OH to C₃H₇COOC₂H₅ is 1:1. The moles of C₃H₇COOC₂H₅ produced from 6.00 moles of C₂H₅OH are:
6.00 mol C₂H₅OH × (1 mol C₃H₇COOC₂H₅/1 mol C₂H₅OH) = 6.00 mol C₃H₇COOC₂H₅
The molar mass of C₃H₇COOC₂H₅ is 116.16 g/mol. The mass corresponding to 6.00 mol is:
6.00 mol × (116.16 g/mol) = 697 g
When the atomic number increases The atomic size will be larger.
A reaction is when two or more pure substances combine with each other to form another identity of pure substances. In general from, it is written as:
A + B → C + D
The substances A and B are the reactants, while the substances C and D are the products. Therefore, in a reaction, the products are found at the end or right side of the reaction.
