Boron (B) has 3 electrons in the outer shell.
Barium (Ba) has 2 electrons.
Phosphorus (P) has 5 electrons.
Manganese (Mn) has 2 electrons.
The answer is C) P (phosphorus).
<span>The gas phosphine (PH3) is used as a fumigant to protect stored grain and other durable produce from pests. Phosphine is generated where it is to be used by adding water to aluminum phosphide or magnesium phosphide. Give formulas for these two phosphides.
</span>Al₂P₃ + 3H₂O → 3PH₃ + Al₂O₃
MgP + H₂O → PH₃ + MgO
Answer:
Dysprosium [Dy]=![[Xe]4f^{10}6s^2](https://tex.z-dn.net/?f=%5BXe%5D4f%5E%7B10%7D6s%5E2)
Americium [Am]=![[Rn]5f^77s^2](https://tex.z-dn.net/?f=%5BRn%5D5f%5E77s%5E2)
Dysprosium is a chemical element with symbol Dy and atomic number of 66. It is a rare earth metal and as it contains partially filled f sub shells, it belongs to f block. Xe is the nearest noble gas and has atomic number of 54.
Americium is a chemical element with symbol Am and atomic number of 95. It is a rare earth metal and as it contains half filled f sub shells, it belongs to f block. Radon is the nearest noble gas and has atomic number of 86.
Answer:
The value of an intensive property may vary with time and its position within the system. Examples of intensive properties include temperature, velocity, mass density, specific volume, and specific energy. An extensive property does not have a value at a point, and its value depends on the extent or size of the system.
Answer:
1.9 × 10² g NaN₃
1.5 g/L
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles of N₂ formed
N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol
We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.
4.2 mol × 28.01 g/mol = 1.2 × 10² g
Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂
The mass ratio of NaN₃ to N₂ is 130.02:84.03.
1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃
Step 4: Calculate the density of N₂
We will use the following expression.
ρ = P × M / R × T
ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L
