Answer:
We need 92.3 grams of sodium azide
Explanation:
Step 1: Data given
Mass of nitrogen gas = 59.6 grams
Molar mass of nitrogen gas = 28.0 g/mol
Molar mass of sodium azide = 65.0 g/mol
Step 2: The balanced equation
2NaN3 → 2Na + 3N2
Step 3: Calculate moles nitrogen gas
Moles N2 = mass N2 / molar mass N2
Moles N2 = 59.6 grams/ 28.0 g/mol
Moles N2 = 2.13 moles
Step 4: Calculate moles NaN3
for 2 moles NaN3 we'll have 2 moles Na and 3 moles N2
For 2.13 moles N2 we need 2/3* 2.13 = 1.42 moles NaN3
Step 5: Calculate mass NaN3
Mass NaN3 = Moles NaN3 * molar mass NaN3
Mass NaN3 = 1.42 moles * 65.0 g/mol
Mass NaN3 = 92.3 grams
We need 92.3 grams of sodium azide
Answer:
0.6 Ω
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12 V
Current (I) = 20 A
Resistance (R) =?
From Ohm's law,
V = IR
Where:
V => is the voltage
I => is the current
R => R is the resistance
With the above formula, we can obtain the resistance as follow:
Voltage (V) = 12 V
Current (I) = 20 A
Resistance (R) =?
V = IR
12 = 20 × R
Divide both side by 20
R = 12 / 20
R = 0.6 Ω
Thus the resistance is 0.6 Ω
Answer:
0.2M NaOh
Explanation:
there are 0.2 mol of NaOH in 8.0 g. (8.0/40) =0.2. Molarity = mol/L = 0.2M.
Delta E = Ef - Ei
E = energy , h = plank constant , v = frequency
h= 6.626 * 10 ^-34 j*s , T = 10 ^ 12 , v = 74 * 10 ^12 Hz , Hz = s^-1
E = ( 6.626 * 10^ -34 j*s) ( 74 * 10 ^ 12 s^ -1 ) = 4.90 * 10 ^ -20 J
Delta E = Ef - Ei
-4.90 * 10 ^ -20 J = -2.18 * 10 ^ -18J ( 1/4 ^2 - 1/x ^2)
0.0225 = 0.0625 - ( 1/x ^ 2)
0.225 - 0.0625 = - 1/ x ^ 2
- 0.0400 = - 1/x ^2 = -1 / - 0.0400 = x^2
25 = x^2
x = 5