Answer:
Carbon or the symbol C has a mass percentage of 42.880% while Oxygen or symbol O has a mass percentage of 57.120%
Answer: Transition from X to Y will have greater energy difference.
Explanation: For studying the energy difference, we require Planck's equation.

where, h = Planck's Constant
c = Speed of light
E = Energy
= Wavelength of particle
From the equation, it is visible that the energy and wavelength follow inverse relation which means that with low wavelength value, energy will be the highest and vice-versa.
As electron A falls from X-energy level to Y-energy level, it releases blue light which has low wavelength value (around 470 nm) which means that it has high energy.
Similarly, Electron B releases red light when it falls from Y-energy level to Z-energy level, which has high wavelength value (around 700 nm), giving it a low energy value.
Energy Difference between X-energy level and Y-energy level will be more.
Answer:
sulfur promotes oxide-reduction reactions.
Explanation:
In stagnant water, some solutes tend to precipitate. When Sulfur precipitate and touch a metal, Sulfur is being reduced and the metal is oxidated. This depends of potential redox of each element.
Au, N, O ( give me brainliest please)
Answer:
- The limiting reactant is lead(II) nitrate.
- 7.20 g of precipitate are formed.
- 1.9 g of the excess reactant remain.
Explanation:
The reaction that takes place is:
- Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
With a percent yield of 87.5%.
To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:
- 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
- 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl
Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.
To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:
- 0.0296 mol Pb(NO₃)₂
*
* 87.5/100 = 7.20 g PbCl₂
- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:
- 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂
Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:
- 0.0259 mol Pb(NO₃)₂
*
= 3.86 g KCl
3.86 g of KCl would react, so the amount remaining would be: