Answer:
the mass of CaO present at equilibrium is, 0.01652g
Explanation:
![K_p = [CO_2]](https://tex.z-dn.net/?f=K_p%20%3D%20%5BCO_2%5D)
= 3.8×10⁻²
Now we have to calculate the moles of CO₂
Using ideal gas equation,
PV =nRT
P = pressure of gas = 3.8×10⁻²
T = temperature of gas = 1000 K
V = volume of gas = 0.638 L
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mole.k
Now we have to calculate the mass of CaO
mass = 2.95 * 10 ⁻⁴ × 56
= 0.01652g
Therefore,
the mass of CaO present at equilibrium is, 0.01652g
The empirical formula of the compound is C. NiF₂.
<em>Step 1</em>. Calculate the <em>moles of each element</em>
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of Ni to F.
Moles of Ni = 9.11 g Ni × (1 mol Ni /(58.69 g Ni) = 0.1552 mol Ni
Moles of F = 5.89 g F × (1 mol F/19.00 g F) = 0.3100 mol F
<em>Step 2</em>. Calculate the <em>molar ratio</em> of the elements
Divide each number by the smallest number of moles
Ni:F = 0.1552:0.3100 = 1:1.997 ≈ 1:2
<em>Step 3</em>: Write the <em>empirical formula</em>
EF = NiF₂
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.
Answer:
The correct answer is 4.58 grams.
Explanation:
Based on the Faraday's law of electrolysis, at the time of electrolysis, the amount of deposited substance is directly equivalent to the concentration of the flow of charge all through the solution. If current, I, is passed for time, t, seconds and w is the concentration of the substance deposited, then w is directly proportional to I*t or w = zIt (Here z refers to the electrochemical equivalent or the amount deposited when 1 C is passed).
For the reaction, n * 96500 C = molar mass
1C = molar mass/n*96500 = Equivalent wt / 96500
w = Equivalent wt / 96500 * I * t
In the given reaction,
Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 + 2H2O, n = 2, the current or I drawn is 350 A, for time, t 12.2 seconds.
Now putting the values in the equation we get,
w = 207.19 / 2 * 96500 * 350 * 12.2 ( The molecular weight of Pb is 207.19 and the equivalent weight of Pb is 207.19 / 2)
w = 4.58 gm.
