Answer:
4180J
Explanation:
(25.0g)(4.184J/g°C)(75°C-35.0°C)
(25.0g)(40.0°C)(4.184J/g°C)
(1.00*10³g°C)(4.184J/g°C) = 4184J
use sig figs:
4180J
A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.
<h3>What does Beer-Lambert law state?</h3>
The Beer-Lambert law states that for a given material sample, path length and concentration of the sample are directly proportional to the absorbance of the light.
A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, we can calculate the molarity of the solution using the following expression.
A = ε × b × c
c = A / ε × b
c = 0.2 / (59 cm⁻¹ M⁻¹) × 1 cm = 0.003 M
where,
- A is the absorbance.
- ε is the path length.
- b is the molar absorptivity coefficient.
- c is the molar concentration.
A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.
Learn more about the Beer-Lambert law here: brainly.com/question/12975133
Answer:
2.645
Explanation:
Rate of diffusion formula:
Sqrt(mass2/mass1)
>>sqrt(14/2)
(Note:Hydrogen must exist in dwiatomic, [H2])
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>
<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
<h2>pH = 8.34</h2>
The answer is that exact locations within either cannot be determined at any given moment in time.
An electron cloud be compared with a spinning airplane propeller in the manner that in both exact location within either cannot be determined at any given moment in time.
In both electron cloud as well as spinning airplane propeller, there is a probability of finding either but exact location can not be determined.
