How many molecules are in 13.5g of sulfur dioxide, so2?

How many molecules are in 13.5g of sulfur dioxide, so2?

2 months ago

Solution 1

Guest Guest #3718
2 months ago

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:

13.5 gramsx\frac{1 mole}{64 grams} = 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

Guest Guest #3719
2 months ago
Answer is: there are 1.27·10²³ molecules of sulfur dioxide.
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

How many valence electrons does a neutral atom of aluminum have? A. 3 B. 10 C. 13 D. More information is needed to figure this out.
Solution 1
A. Valence electrons are the electrons on the outermost energy level. If you write out the electron configuration you add all of the electrons on the 3rd energy level. 
Which substance is acting as the Brønsted-Lowry acid in the following chemical reaction? NH4 + OH- yields NH3 + H2O A. NH4+ B. OH- C. NH3 D. H2O
Solution 1
The answer to this question is A. NH4+. Its conjugate base is NH3 so NH4 donates a proton to OH- to become NH3 in the equilibrium reaction.
Solution 2
The  the Brønsted-Lowry acid donates H⁻.
In this reaction Particle that  loose H⁺ is A. NH4⁺ ion.
I think the answer is B, but I just looked it up, so yeah. I need a real answer, no guessing The speed of sound is typically measured in: A. cm/sec B. m/sec C. km/sec D. mi/sec E. mi/min ps- why is there no science category???
Solution 1
The answer is B lol you had it right
Calculate the pressure in atm, if 0.00825 moles occupies 174 mL at -15 celcius.
Solution 1
Answer is: the pressure is 1.12 atm.
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.

How many moles of 18L of NH3 at 30 celcius and 912 mmHg? how many grams?
Solution 1
Answer is: mass of ammonia is 14.76 grams and amount is 0.87 moles.
) = 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.
 = 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.

Identify the atom represented by this electron configuration and elaborate on its bonding properties. 1s22s22p63s23p64s23d5 A) .Chromium is a transition with a half full-valence orbital and can form up to six bonds. B) Antimony is a metalloid that can form five covalent bonds or three metallic bonds. C) Manganese is a transition metal that can form up to five stable ionic bonds D) Bromine is a non-metal that forms one ionic bond or a variable number of covalent bonds.
Solution 1
            The given electronic configuration represents Manganese.

                   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
                                                  (4 - 1)d


So tracking 4rth period, we found Mn with given electronic configuration.
        "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.
How are solids and liquids different than gases in terms of volume changes
Solution 1
They are different substances because in the weight and size, they are different.
in weight gas weighs nothing but every solid is different.
same with liquids.
How many moles and grams of NaOH are needed to dissolve in water to make 4 liters of a 2.0M solution?
Solution 1
The answers are 8.0 moles and 320 grams.
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
Please help!!! How would you calculate the % yield of hydrogen if 40.0 grams of magnesium react with an excess of nitric acid producing 1.70 grams of hydrogen gas. Mg + 2 HNO3 → Mg(NO3)2 + H2
Solution 1
Answer is: yield of hydrogen is 51.5%.
Balanced chemical reaction: Mg + 2HNO
₃ → 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.
Solution 2

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.


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:

if 392cm2 of helium at 32C and 925 torr are cooled to -15C and the pressure is reduced to 775 torr, calculate the new volume the gas will occupy
Solution 1
The answer is 553 cm³.
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