Solution 1
Taking into account the definition of avogadro's number, 0.21 moles of sulfur dioxide contain 1.26482×10²³ molecules.
Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.
First you must determine the number of moles that 13.5 g of sulfur dioxide contains. For that, I use the molar mass of the compound, which is defined as the amount of mass that a substance contains in one mole.
In this case, the molar mass of sulfur dioxide is 64 g/mole. So the number of moles that 13.5 grams of the compound contain can be calculated as:
0.21 moles
Then you can apply the following rule of three: if 1 mole of sulfur dioxide contains 6.023×10²³ molecules, then 0.21 moles contain how many molecules of sulfur dioxide?
amount of molecules of sulfur dioxide= (6.023×10²³ molecules× 0.21 mole)÷ 1 mole
amount of molecules of sulfur dioxide=1.26482×10²³ molecules
Finally, 0.21 moles of sulfur dioxide contain 1.26482×10²³ molecules.
Learn more about Avogadro's Number:
Solution 2
m(SO₂) = 13.5 g.
n(SO₂) = m(SO₂) ÷ M(SO₂).
n(SO₂) = 13.5 g ÷ 64 g/mol.
n(SO₂) = 0.21 mol.
N(SO₂) = n(SO₂) ·Na.
N(SO₂) = 0.21 mol · 6.022·10²³ 1/mol.
N(SO₂) = 1.27·10²³.
n - amount of substance.
M - molar mass of substance.
Na - Avogadro number.
📚 Related Questions
In this reaction Particle that loose H⁺ is A. NH4⁺ ion.
n(O₂) = 0.00825 mol, amount of substance.
V = 174 mL ÷ 1000 mL/L = 0.174 L, volume of gas.
T = 15°C = 288.15 K; temperature.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
p = n·R·T / V.
p = 0.00825 mol · 0.08206 L·atm/mol·K · 288.15 K / 0.174 L.
p = 1.12 atm.
V(NH₃) = 18 L.
T = 30°C = 303.15 K.
p = 912 mmHg ÷ 760 mmHg/atm= 1.2 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
n = p·V / R·T.
n(NH₃) = 1.2 atm ·18 L / 0.08206 L·atm/mol·K · 303.15 K.
n(NH₃) = 0.87 mol.
m(NH₃) = n(NH₃) · M(NH₃).
m(NH₃) = 0.87 mol · 17 g/mol.
m(NH₃) = 14.76 g.
The given electronic configuration represents Manganese.
Explanation:
The given electronic configuration is....
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁵
As the valence shell is 4 in electronic configuration so it is present in 4rth period in periodic table. And according to definition of transition elements, transition elements has partially filled d-orbital. i.e.
(n - 1)d
As, n = 4
So,
(4 - 1)d
3d
So tracking 4rth period, we found Mn with given electronic configuration.
Also,
"Manganese is a transition metal that can form up to five stable ionic bonds". It means that the five electrons present in d-orbital are transferred and Mn converts into Mn⁺⁵.
in weight gas weighs nothing but every solid is different.
same with liquids.
Solution:
Since we are given the concentration of the solution expressed in molarity and the volume of solution, we can use the equation for molarity defined as the number of moles of solute divided by the volume of the solution in liters:
molarity = moles of solute / volume of solution in liters
We rearrange the expression to solve for the number of moles of the solute:
moles of solute = molarity * volume of solution in liters
Substituting the given values, we have
moles of NaOH = (2.0 mol / 1L) * 4L = 8.0 mol
To calculate for the mass of NaOH solute, we will use the molar mass of NaOH that is equal to 40.0 g/mol:
mass of NaOH = 8.0 mol * (40.0g / 1mol) = 320 grams
Balanced chemical reaction: Mg + 2HNO₃ → Mg(NO₃)₂ + H₂.
m(Mg) = 40.0g.
n(Mg) = m(Mg) ÷ M(Mg).
n(Mg) = 40 g ÷ 24.3 g/mol.
n(Mg) = 1.65 mol.
From chemical reaction: n(Mg) : n(H₂) = 1 : 1.
n(H₂) = n(Mg) = 1.65 mol.
m(H₂) = 1.65 mol · 2 g/mol.
m(H₂) = 3.3 g.
yield = 1.70 g ÷ 3.3 g · 100% = 51.5%.
n - amount of substance.
M - molar mass.
The percentage yield of Hydrogen is 51.1%
We'll begin by calculating the mass of Mg that reacted and the mass of H₂ produced from the balanced equation.
Mg + 2HNO₃ → Mg(NO₃)₂ + H₂
Molar mass of Mg = 24 g/mol
Mass of Mg from the balanced equation = 1 × 24 = 24 g
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ from the balanced equation = 1 × 2 = 2 g
From the balanced equation above,
24 g of Mg reacted to produce 2 g of H₂.
Next, we shall determine the theoretical yield of H₂.
From the balanced equation above,
24 g of Mg reacted to produce 2 g of H₂.
Therefore,
40 g of Mg will react to produce = (40 × 2) / 24 = 3.33 g of H₂.
Thus the theoretical yield of H₂ is 3.33 g
Finally, we shall determine the percentage yield of H₂.
- Actual yield = 1.70 g
- Theoretical yield = 3.33 g
- Percentage yield =?
Percentage yield = (Actual / Theoretical) × 100
Percentage yield = (1.70 / 3.33) × 100
Percentage yield of H₂ = 51.1%
Learn more about percentage yield:
Solution:
Assuming ideal behavior of the helium gas, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the final conditions.
We rearrange the gas law equation to calculate for the new volume V2:
V2 = P1V1T2 / P2T1
We can now substitute the values to the expression. Therefore,
V2 = (925torr * 392cm3 * 258.15K) / (305.15K * 775torr)
V2 = 396 cm³ or 396 553 mL