Answer:
sure
Explanation:
The substance formed after heating the mixture of that of Rahul is caleed a compound. Whereas, Manav's mixture still remains in its current stae that is a heterogeneous mixture.
The compound formed is in black in color whereas the mixture is a mix of brownish-red and yellow.
The compound is a homogeneous mixture whereas the mixture is a heterogenous mixture because of its uneven distribution.
The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV
Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules
From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s
Answer:
The pH of the solution is 8.
Explanation:
To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:
1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:
pOH = –Log [OH-]
pOH = 6
6 = –Log [OH-]
–6 = Log [OH-]
[OH-] = Antilog (–6)
[OH-] = 1x10^–6 mol/L
2. The pH of the solution can be obtained as follow:
pH + pOH = 14
pOH = 6
pH + 6 = 14
pH = 14 – 6
pH = 8.
From the calculations made above,
[OH-] = 1x10^–6 mol/L
pH = 8.
Therefore, the correct answer is:
The pH of the solution is 8
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
one in a 2s orbital
Explanation:
Because of the peak near the nucleus in the 2s curve there is a higher probability of finding a 2s within 4 Å of the nucleus. In a multi-electron atom an electron in a 2s orbital will have a lower energy than one in a 2p orbital