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gizmo_the_mogwai [7]

3 years ago

6

When methane is burned with oxygen, the products are carbon dioxide and water. if you produce 9 grams of water and 11 grams of c

arbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction? 4 31 20 40 none of the above?

1 answer:

Alina [70]3 years ago

3 0

4 grams of methane is <span>burned with oxygen,. Hope this helped</span>

You might be interested in

The atomic number is the number of protons in the atom of the element

ipn [44]

Answer:

True

Explanation:

If you look closley at the nucleus, you don't count the neutrons just the prtons which then effect the electrons.

Good luck :)

5 0

3 years ago

What is the empirical formula of a compound that is 7.74% H and 92.26% C? What is the molecular formula if the molar mass is 78.

Minchanka [31]

Answer:

For all these questions, we want to find the empirical and molecular formulae of various compounds given their percent composition and molar mass. The technique used to answer one of the questions can accordingly be applied to all of them.

Approaching the first question, we treat the percentages of each element as the mass of that element in a 100 g compound (as the percentages add up to 100%). So, our 100 g compound comprises 7.74 g H and 92.26 g C.

Next, we convert these mass quantities into moles. Divide the mass of each element by its molar mass:

7.74 g H/1.00794 g/mol = 7.679 mol H

92.26 g C/12.0107 g/mol = 7.681 mol C.

Then, we look for the molar quantity that's the smallest ("smaller," in this case, since there are only two), and we divide all the molar quantities by the smallest one. Here, it's a very close call, but the number of moles of H is slightly smaller than that of C. So, we divide each molar quantity by the number of moles of H:

7.679 mol H/7.679 mol H = 1

7.681 mol C/7.679 mol H ≈ 1 C/H (the value is actually slightly larger than 1, but we can treat it as 1 for our purposes).

The quotients we calculated represent the subscripts of our compound's empirical formula, which should provide the most simplified whole number ratio of the elements. So the empirical formula of our compound is C₁H₁, or just CH.

Here, it just so happens that we obtained whole number quotients. If we end up with a quotient that isn't a whole number (e.g., 1.5), we would multiply all the quotients by a common number that <em>would </em>give us the most simplified whole number ratio (so, if we had gotten 1 and 1.5, we'd multiply both by 2, and the empirical formula would have subscripts 2 and 3).

To find the molecular formula (the actual formula of our compound), we use the molar mass of the compound, 78.1134 g/mol. The molar mass of our "empirical compound," CH, is 13.0186 g/mol. Since our empirical formula represents the most simplified molar ratio of the elements, the molar masses of our "empirical compound" and the actual compound should be multiples of one another. We divide 78.1134 g/mol by 13.0176 g/mol and obtain 6. The subscripts in our molecular formula are equal to the subscripts in our empirical formula multiplied by 6.

Thus, our molecular formula is C₆H₆.

---

As mentioned before, all the questions here can be answered following the procedure used to answer the first question above. In any case, I've provided the empirical and molecular formulae for the remaining questions below for your reference.

2. Empirical formula: C₁₃H₁₂O; molecular formula: C₁₃H₁₂O

3. Empirical formula: CH; molecular formula: C₈H₈

4. Empirical formula: C₂HCl; molecular formula: C₆H₃Cl₃

5. Empirical formula: Cl₄K₂Pt; molecular formula: Cl₄K₂Pt

6. Empirical formula: C₂H₄Cl; molecular formula: C₄H₈Cl₂

6 0

2 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

<u>Answer:</u>

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

<u>Explanation:</u>

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

<u>Cell having 1st and 2nd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

<u>Cell having 1st and 3rd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

<u>Cell having 1st and 4th half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

<u>Cell having 2nd and 3rd half reactions:</u>

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

<u>Cell having 3rd and 4th half reactions:</u>

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

3 years ago

H+(aq) + OH–(aq) → H2O(l) + 55.8 kJ In this reaction there is conservation of

pentagon [3]

In the equation given above, there is conservation of MASS, CHARGE AND ENERGY.
These three parameters are usually conserved during the course of chemical reactions. When any of these parameter experience a reduction during the course of chemical reaction, such loss is always gained by other elements involved in the same reaction, so that at the end of the day, they are not considered as lost.

4 0

3 years ago

What would the molecule CH₄ be classified as?

larisa [96]

Answer:

Alkane

Explanation:

Definition of Alkane "any of the series of saturated hydrocarbons including methane, ethane, propane, and higher members. (google dictionary)"

CH4 is methane.

6 0

3 years ago