Answer:
3s and 3p
Explanation:
From the question given above, the following data were obtained:
Electronic configuration =>
1s² 2s²2p⁶ 3s²3p⁴
Location of valence electron =?
From the electronic configuration given above, we can see clearly that the atom has three (3) shells.
Valence electron(s) are located at the outer most shell of an atom.
The outer most shell of the atom above is 3s and 3p.
Therefore, 3s and 3p will contain the valence electron(s)
Answer:
YES
Explanation:
WE use science cause it is all aaprindhs and science us everywhere .
Answer:
a) carboxylic acid
b) amine
c) ester
d) aldehyde
e) alkene
f) ketone
Explanation:
Most of them are straight forward.
Let's say say there are n1 mols of helium in the first balloon and n2 mols of nitrogen in the second one, which are equivalent to m1 grams of helium and m2 grams of nitrogen.
The molar mass of hydrogen is thus M1=m1/n1, same for nitrogen M2=m2/n2 hence the ratio of their masses is m1/m2=(M1n1)/(M2n2). Since both gases are rather similar, we can assume that n1~n2 hence m1/m2=M1/M2
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
