Answer:
you need to send us the figure
Explanation:
Answer:
The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Explanation:
Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg
The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)
where u = initial rms speed of the gas = 850 m/s
Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J
Then, 3600 J is added to the gas,
New kinetic energy of the gas = 4335 + 3600 = 7935 J
New kinetic energy of Helium atoms = (1/2)(m)(v²)
where v = final rms speed of the gas = ?
7935 = (1/2)(0.012)(v²)
v² = (7935×2)/0.012
v² = 1,322,500
v = 1150 m/s
Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Hope this Helps!!!
Answer:
The structure can be found on the attached documents
Explanation:
Answer:
3.02× 10²⁴ atoms
Explanation:
Given data:
Number of nitrogen atoms = ?
Number of moles of N₂O = 2.51 mol
Solution:
1 mole contain 2 mole of nitrogen atoms.
2.51 × 2 = 5.02 mol
According to Avogadro number,
1 mole = 6.022 × 10²³ atoms
5.02 mol × 6.022 × 10²³ atoms / 1 mol
30.2 × 10²³ atoms
3.02× 10²⁴ atoms
First calculate the moles of N2 and H2 reacted.
moles N2 = 27.7 g / (28 g/mol) = 0.9893 mol
moles H2 = 4.45 g / (2 g/mol) = 2.225 mol
We can see that N2 is the limiting reactant, therefore we
base our calculation from that.
Calculating for mass of N2H4 formed:
mass N2H4 = 0.9893 mol N2 * (1 mole N2H4 / 1 mole N2) * 32
g / mol * 0.775
<span>mass N2H4 = 24.53 grams</span>