Answer:
Explanation:
To convert from moles to grams, the molar mass must be used.
1. Find Molar Mass
The compound is sodium sulfide: Na₂S
First, find the molar masses of the individual elements in the compound: sodium (Na) and sulfur (S).
- Na: 22.9897693 g/mol
- S: 32.07 g/mol
There are 2 atoms of sodium, denoted by the subscript after Na. Multiply the molar mass of sodium by 2 and add sulfur's molar mass.
- Na₂S: 2(22.9897693 g/mol)+(32.07 g/mol)=78.0495386 g/mol
This number tells us the grams of sodium sulfide in 1 mole.
2. Calculate Grams
Use the number as a ratio.
Multiply by the given number of moles, 3.46.
The moles of sodium sulfide will cancel.
3. Round
The original measurement of grams, 3.46, has 3 significant figures. We must round our answer to 3 sig figs.
For the answer we calculated, that is the ones place.
The 0 in the tenth place tells us to leave the 0 in the ones place.
There are about <u>270 grams of sodium sulfide</u> in 3.46 moles.
Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams
The number of mole will be 65.81 mole.
An ideal gas would be one for which both the overall volume of the molecules and even the forces that exist between them are so negligible as to have no influence on the behavior of something like the gas.
Number of ideal gas can be calculated by using the formula:
PV = nRT
where, p is pressure, n is number of mole, R is gas constant and T is temperature.
Given data:
V= 1750 = 1750 L
P = 125,000 p = 1.2 atm
R = 0.082 L /mole kelvin
T = 273+127 = 400 K
Now, put the value of given data in above equation.
1.23atm x 1750L = n x 0.0820atm x Liter/ mole x kelvin x 400K
n = 65.81 mole.
Therefore, the number of mole will be 65.81 mole
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Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
Answer:
voltage source
Explanation:
establishes an electric potential difference across the two ends of the external circuit and thus causes the charge to flow. The voltage source is the numerical value of this electric potential difference.