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
Br
Se
As
Ge
Ga
Rb
Elements in the top right corner of the periodic table have the highest electronegativity. Elements on the right side have a higher electronegativity than those on the left, same with the ones on the top in comparison to those on the bottom.
Answer:
Rubidium Acetate
Explanation:
Rb is Rubidium C2H3O2 is acetate
Answer:
1,300,000,000,000
Explanation:
1.3 x 10^12
We want to convert this from scientific notation.
Tip: in scientific notation the exponent tells you how many place you move the decimal point over. If the exponent is negative you move the decimal point to the left. Ex. For, 4.1 x 10^-8, we would move the decimal point over 8 times to the left to get .00000041. When the exponent is positive we move over to the right. Ex. For, 7.6 x 10^7 we would move the decimal point over 7 times to the right to get 76,000,000
So to convert 1.3 x 10^12 we simply move over the decimal point over 12 times to the right.
1.3 x 10^12 ------> 1,300,000,000,000
Our answer is 1,300,000,000,000
Yes. It does contain Ions.
