2 Cr + 3 Cl2 ------> 2CrCl3
A ground state electron configuration follows the Aufbau Principle that states that electrons should be filled up in orbitals in increasing energy. In the given sequences, the right configuration is
<span>1s2 2s2 2p6 3s2 3p6 4s2 3d8.
2) the possible confirmation that follows Aufbau's principle is
D. </span><span>-[Kr] 5s24d105p3</span>
Answer: A mass of 124457.96 g ammonia is produced by reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm.
Explanation:
Given: Volume = 450 L
Temperature = 450 K
Pressure = 300 atm
Using ideal gas equation, moles of nitrogen are calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = tempertaure
Substitute values into the above formula as follows.
According to the given equation, 1 mole of nitrogen forms 2 moles of ammonia. So, moles of ammonia formed by 3654.08 moles of nitrogen is as follows.
As moles is the mass of substance divided by its molar mass. So, mass of ammonia (molar mass = 17.03 g/mol) is as follows.
Thus, we can conclude that a mass of 124457.96 g ammonia is produced by reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm.
The decay mode of cesium-137 is beta decay. This means that the cesium-137 decays into a beta particle and a nuclide with the same mass number, but with a charge number that is 1 more than that of cesium.
Therefore, this means Cs-137 decays into an electron and Barium-137, meaning the answer is choice 1.
Answer:
39.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction
Ba(OH)₂(aq) + 2 CH₃COOH(aq) ⟶ Ba(CH₃COO)₂(aq) + 2 H₂O(l)
Step 2: Calculate the moles corresponding to 2.78 g of CH₃COOH
The molar mass of CH₃COOH is 60.05 g/mol.
2.78 g × 1 mol/60.05 g = 0.0463 mol
Step 3: Calculate the moles of Ba(OH)₂ needed to react with 0.0463 moles of CH₃COOH
The molar ratio of Ba(OH)₂ to CH₃COOH is 1:2. The moles of Ba(OH)₂ needed are 1/2 × 0.0463 mol = 0.0232 mol.
Step 4: Calculate the volume of 0.586 M solution that contains 0.0232 moles of Ba(OH)₂
0.0232 mol × 1 L/0.586 mol = 0.0396 L = 39.6 mL
