Answer:
The correct approach is Option B (Peer Review).
Explanation:
- Rather made reference to someone as a scientific peer-review, it encourages the specialist who has not been essential to the study team to analyze the study objectively and pointed out everyone's mistakes. It serves as major self-regulation for scholars and aims to make the publishing process somewhat credible. Hence, the solution to this issue is Peer Examination.
- Funding organizations rarely have the capabilities to recognize out mistakes, whereas definitive analysis is a method of study that helps to make a definitive statement. The gathering of data is simply a process of scientific study.
Other approaches do not apply to the example mentioned. Although the one mentioned is right.
The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
learn more about Ions:
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Answer:
713 nm. It is not visible with the naked eye.
Explanation:
Step 1: Given data
- Energy of light (E): 2.79 × 10⁻¹⁹ J
- Planck's constant (h): 6.63 × 10⁻³⁴ J.s
- Speed of light (c): 3.00 × 10⁸ m/s
Step 2: Calculate the wavelength of the light
We will use the Planck-Einstein equation.
E = h × c / λ
λ = h × c / E
λ = 6.63 × 10⁻³⁴ J.s × 3.00 × 10⁸ m/s / 2.79 × 10⁻¹⁹ J
λ = 7.13 × 10⁻⁷ m
Step 3: Convert "λ" to nm
We will use the relationship 1 m = 10⁹ nm.
7.13 × 10⁻⁷ m × (10⁹ nm/1 m) = 713 nm
This light is not in the 400-700 nm interval so it is not visible with the naked eye.
It is definitely a molecule
