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Anni [7]
3 years ago
13

Given the following equation, what is the correct form of the conversion factor needed to convert the number of moles O2 to the

number of moles of Fe reacted?
4Fe + 3O2 --> 2Fe2O3

4 mol Fe
3 mol O2


2 mol Fe2O3
4 mol Fe


4 mol Fe
2 mol Fe2O3


3 mol O2
2 mol Fe2O3
Chemistry
2 answers:
julsineya [31]3 years ago
8 0

Answer:

The correct answer is the first option: 4 mol Fe , 3 mol O2

Explanation:

Hello!

Let's solve this!

The reaction is:

4Fe + 3O2 ---> 2Fe2O3

This reads that every 4 moles of iron consumed, we need 3 moles of oxygen and when they react, we get 2 moles of Fe2O3.

Let's analyze each of the options:

4 mol Fe

3 mol O2: this is correct since it tells us how much iron and how much oxygen we need.

2 mol Fe2O3

4 mol Fe: it is not correct because it does not tell us anything about oxygen.

4 mol Fe

2 mol Fe2O3: it is not correct because it does not tell us anything about oxygen.

3 mol O2

2 mol Fe2O3: it is not correct because, although it says of oxygen, it does not relate it to iron.

We conclude that the correct answer is the first option: 4 mol Fe

3 mol O2

jek_recluse [69]3 years ago
4 0

Answer:

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Explanation:

Step 1: Data given

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate the mol ratio

For 3 moles O2 we'll have 4 moles Fe

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Option b says we have 2 mole Fe2O3 for each 4 moles Fe

This doesnt say anything about O2. So doesn't apply for this question.

Option C says we have 4 moles of Fe for each 2 moles Fe2O3

This is the same as option B, so doesn't apply for this question.

Option D says for each 3 moles of O2 we have 2 Fe2O3

This is true, but doesn't say anything about Fe so doesn't apply here.

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What factors affect the dynamic state of equilibrium in a chemical reaction and how?
yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

  • External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

  • Changes in pressure due to volume changes.

Catalysts do not influence the value of Q. See explanation.

Explanation:

\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

Similar to the rate constant, the equilibrium constant K_c depends only on:

  • \Delta G the standard Gibbs energy change of the reaction, and
  • T the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

  • Changes in concentration influence the number of particles per unit space.
  • Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

  • Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

8 0
3 years ago
Which pair of aqueous solutions can create a buffer solution if present in the appropriate concentrations?.
Julli [10]

HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.

<h3>What are buffer solutions and how do they differ?</h3>
  • The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
  • Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
  • For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
  • When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
  • A buffer made of a weak acid and its salt is an example.
  • It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.

learn more about buffer solutions here

<u>brainly.com/question/8676275</u>

#SPJ4

8 0
1 year ago
How many atoms are there in 8.88 g Si?
Mariana [72]

Answer:

\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

<h3>1. Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

  • Si:  28.085 g/mol

We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

\frac { 28.085 \ g \  Si}{1 \ mol \ Si}

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

8.88 \ g \ Si *\frac { 28.085 \ g \  Si}{1 \ mol \ Si}

Flip the fraction so the units of grams of silicon cancel.

8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \  Si}

8.88  *\frac{1 \ mol \ Si} { 28.085 }

\frac {8.88} { 28.085 } \ mol \ Si

0.316183015845 \ mol \ Si

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}

Multiply by the number of moles we calculated.

0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}

The units of moles of silicon cancel.

0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}

0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}

1.90405412 \times 10^{23} \ atoms \ Si

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

1.90 \times 10^{23} \ atoms \ Si

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

6 0
3 years ago
Be sure to answer all parts. Nitric oxide, NO·, is a radical thought to cause ozone destruction by a mechanism similar to that o
SIZIF [17.4K]

Answer:

See image attached and explanation

Explanation:

The stratospheric ozone layer is very important in absorbing high-energy ultraviolet radiation that is harmful to living systems on earth. The concentration of ozone in the stratosphere is determined by both thermal and photochemical pathways for its decomposition. Nitric oxide, NO, is a trace constituent in the stratosphere that reacts with ozone to form nitrogen dioxide, NO2, and the diatomic oxygen molecule. The nitrogen-oxygen bond in NO2 is relatively weak. When an NO2 molecule encounters an oxygen atom, it transfers an oxygen, forming O2 and NO. The chemical reactions involved are formations of NO2 following by reaction of NO2 with atomic oxygen for form NO and O2. The sum of both reactions show that the overall reaction is simply the reaction of ozone with atomic oxygen to form two molecules of molecular oxygen. Hence, NO only serves as a catalyst, it does not undergo a permanent change itself.

6 0
3 years ago
Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere
Colt1911 [192]
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
            = (2x10^-16)/(1x10^-7)^2
             = 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4 
when POH = -㏒[OH-]
            4  = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
          = 2x10^-16 / (1x10^-4)^2
          = 2x10^-8 M
c) at PH= 14 
when POH = 14-PH
                   = 14 - 14 
                   = 0
when POH = -㏒[OH]
              0 = - ㏒[OH]
∴[OH] = 1 m 
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
          = (2x10^-16) / 1^2
          = 2x10^-16 M


8 0
3 years ago
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