Answer : The initial quantity of sodium metal used is 17.25 grams.
Solution : Given,
Volume of
gas = 8.40 L
Molar mass of Na metal = 23 g/mole
The Net balanced chemical reaction is,

At STP, 22.4 L of volume is occupied by 1 mole of
gas
so, 8.40 L of volume is occupied by =
= 0.375 moles of
gas
Now from the above reaction, we conclude that
1 mole of
gas produced by the 2 moles of Na metal
0.375 moles of
gas produced =
of Na metal
The quantity of Na metal used = Moles of Na metal × Molar mass of Na metal = 0.75 moles × 23 g/mole = 17.25 grams
Therefore, the initial quantity of sodium metal used is 17.25 grams.
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The N from NH3 forms a coordinate
covalent bond with the metal iron as in this case. The iron itself is at the "center" of the ligate complex ion. Since the net ionic charge is at 2+, it can be said that there is one pair of unpaired electrons for <span>[fe(nh3)6]2+.</span>
Answer:
Mass of helium = 4426.9524 g
Mass of hydrogen gas = 2213.4762 g
Explanation:
Pressure = 135 atm
Temperature = 24 °C
Volume = 200 L
Number of moles = ?
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (24 + 273.15) K = 297.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
135 atm × 200 L = n × 0.0821 L.atm/K.mol × 297.15 K
⇒n = 1106.7381 moles
<u>For helium gas:</u>
Molar mass = 4 g/mol
The formula for the calculation of moles is shown below:
Thus,

<u>For hydrogen gas:</u>
Molar mass = 2 g/mol
The formula for the calculation of moles is shown below:
Thus,

Answer:
Here is the solution..Hope it helps:)