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Vikki [24]
3 years ago
6

The atomic mass of sulfur is 32.1 amu, and the atomic mass of oxygen is 16.0 amu. To the nearest tenth of a percent, what is the

percent by mass of sulfur in sulfur trioxide (SO3)?
Chemistry
1 answer:
Zielflug [23.3K]3 years ago
6 0

Answer:

\%\ Composition\ of\ sulfur=40.1\ \%

Explanation:

Percent composition is percentage by the mass of element present in the compound.

Given , Mass of sulfur= 32.1 amu

Mass of oxygen = 16.0 amu

Mass of sulfur trioxide SO_3 = 32.1 amu + 3*16.0 amu = 80.1 amu

\%\ Composition\ of\ sulfur=\frac{Mass_{sulfur}}{Mass_{SO_3}}\times 100

\%\ Composition\ of\ sulfur=\frac{32.1\ amu}{80.1\ amu}\times 100=40.1\ \%

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When the following equation is balanced using the smallest possible integers, what is the number in front of the substance in bo
weeeeeb [17]

The balanced equation :

8Al + 3Fe₃O₄→ 4Al₂O₃ + 9Fe

You just have to look for which element is in bold because the question is not clear

<h3>Further explanation</h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

Reaction

Al + Fe₃O₄→ Al₂O₃ + Fe

give a coefficient

aAl + Fe₃O₄→ bAl₂O₃ + cFe

O, left=4, right=3b⇒3b=4⇒b=4/3

Al, left=a, right=2b⇒a=2b⇒a=2.4/3⇒a=8/3

Fe, left=3, right=c⇒c=3

The equation becomes :

8/3Al + Fe₃O₄→ 4/3Al₂O₃ + 3Fe x3

8Al + 3Fe₃O₄→ 4Al₂O₃ + 9Fe

7 0
3 years ago
Snape grows tired of these conceptual questions and thinks it's time for a problem. What is the retention factor if the distance
MAVERICK [17]

Answer:

The Retention factor (rf) value is = 0.2

Explanation:

  • Retention factor (Rf) is factor used substances that could be separated using Chromatography. Retention factor determines how fast the component can move on the chromatogram (stationary phase) after elution. Elution occurs when mobile phase (solvent) moves across the stationary phase when the solute has been spotted on the origin.
  • Retention factor (Rf) ranges from value between 0 and  1. The closer the value to 1, the faster it can move upon elution. Rf can be calculated.
  • Rf value = distance moved by the solute / distance moved by the solvent

             = 0.40cm / 2.00cm

             = 0.2

6 0
3 years ago
The reaction between ethyne (acetylene, C 2 H 2 ) and hydrogen. The product is ethane (C 2 H 6 ). Which is the limiting reactant
serious [3.7K]

Answer:

Three possible cases:

- If amount are equal for each reactant (for example 1 mol each), the limiting is the hydrogen and the excess reagent is the acetylene.

- When moles of H₂ are greater than C₂H₂

The acetylene is the limiting reagent so the H₂ is the excess

-  When moles of C₂H₂ are greater than H₂

For this case, H₂ is the limiting reactant and the excess is the C₂H₂

Explanation:

First of all we determine the reaction:

Reactants, acetylene and hydrogen

Products are ethane

Then, the balanced reaction is: C₂H₂ + 2H₂ → C₂H₆

1 mol of acetylene reacts with 2 moles of hydrogen ir order to produce 1 mol of ethane.

If amount are equal for each reactant, the limiting is the hydrogen,

For example, 1 mol each

For 1 mol of acetylene I need 2 moles of H₂. I've only got 1 mol, so I do not have enough H₂. The excess reagent is the acetylene.

- When moles of H₂ are greater than C₂H₂

For example, 3 moles of H₂ and 0.5 mol of C₂H₂

2moles of H₂ need 1 mol of C₂H₂ for the reaction

Then 3 moles of H₂ will need (3 . 1) / 2 = 1.5 moles

We have 0.5 moles, so the acetylene is the limiting reagent, again.

- When moles of C₂H₂ are greater than H₂

For example 1 mol of C₂H₂ and 0.001 moles of H₂

If I have 1 mol of C₂H₂, I definetly need the double of moles of hydrogen, so in this case, H₂ is the limiting reactant and the excess is the C₂H₂

If we have 1 mol of H₂ and 0.5 mol of C₂H₂, notice that moles of acetylene are lower than hydrogen

1 mol of C₂H₂ needs 2 moles of H₂

So 0.5 moles of C₂H₂ will need 1 mol of H₂ (it's ok because we have 1 mol)

2 moles of H₂ need 1 mol of C₂H₂ for reaction

Then, 1 mol of H₂ will need 0.5 moles of C₂H₂ (it's ok because we have that amount)

In this case, there is no excess neither limiting. That's why we can choose any of them to determine the moles (or mass) for the product

7 0
3 years ago
Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s&gt; CO (s) + Zn2+ (aq)
Simora [160]

<u>Answer:</u> The K_{eq} of the reaction is 1.73\times 10^{16}

<u>Explanation:</u>

For the given half reactions:

Oxidation half reaction: Zn(s)\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V

Reduction half reaction: Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V

Net reaction: Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.28-(-0.76)=0.48V

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

\Delta G^o=-nfE^o_{cell}

and,

\Delta G^o=-RT\ln K_{eq}

Equating these two equations, we get:

nfE^o_{cell}=RT\ln K_{eq}

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = 25^oC=[273+25]=298K

K_{eq} = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}

Hence, the K_{eq} of the reaction is 1.73\times 10^{16}

8 0
3 years ago
Why do planets in our solar system stay in their positions
Anon25 [30]
This is not chemistry but it's A, off top.

4 0
4 years ago
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