CHEMISTRY COLLEGE

(b) Liquid hexane burns in oxygen gas to form carbon dioxide gas and water vapor.
True/False

Answers

Answer 1

Answer:

Answer:

True

Explanation:

Liquid hexane burns in oxygen gas to form carbon dioxide gas and water vapor.TRUE.

This is the complete combustion reaction of hexane, which proceeds according to the following equation.

C₆H₁₄(l) + 9.5 O₂(g) → 6 CO₂(g) + 7 H₂O(g)

If the combustion were incomplete, instead of carbon dioxide, the product would be carbon monoxide or carbon.

Related Questions

Are these statements true or false? Correct the statements that are false.a ..An atom is larger than the cell of a living thing.b.. Different kinds of elements have different kinds of atoms.c..The protons, neutrons and electrons are different for different kinds of atoms.d... An atom always has the same number of protons and electrons.e... An atom always has the same number of protons and neutrons.f.... Atoms have no mass because they are very small.Q2 .Compare the Rutherford's model and the Bohr's model of the atom. State one similarity and one difference between them. Q3..what observation from rutherford's gold foil experiment made him conclude that an atom has a tiny but dense nucleus that is positively charged?Q4.. Explain why the nucleus of an atom is positively charged, while the atom is electrically neutral.Q5...The symbols and atomic numbers of three elements are as follows:Ne Atomic number 10A/ Atomic number 13K Atomic number 19a.. identify each element from its symbol.b.. How many protons and electrons does an atom of each element have?c.. Draw the electronic structure of the atom of each element.
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A chemist prepares a solution of magnesium fluoride MgF2 by measuring out 0.00598μmol of magnesium fluoride into a 50.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's magnesium fluoride solution. Round your answer to 2 significant digits.
Find the weighted average of these values. Col1 Value 5.00 6.00 7.00Col2 Weight 75.0 % 15.0 % 10.0 %
You have 0.500 L of an 0.250 M acetate buffer solution (i.e. [HC₂H₃O₂] + [C₂H₃O₂⁻] = 0.250 M) at pH 3.50. How many mL of 1.000 M NaOH must you add in order to change the pH to 5.07? Acetic acid has a pKa of 4.74.

The element oxygen has valence electrons

Answers

Answer:

it’s electron configuration is 1s^2 2s^2 2p^4. To determine valence electrons, add the outermost s and p orbitals. In an oxygen atom, 8 electrons are present. Electron present in the first shell (n=1) 2n^2=2 (1)^2=2 (1)=2.

Be sure to answer all parts. The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates: 4 NH3(g) +3 O2 (g) ⇌ 2 N2(g) + 6 H2O(g) When 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00−L container at a certain temperature, the N2 concentration at equilibrium is 1.96 × 10−3 M. Calculate Kc.

Answers

Answer: The value of $K_c$ for the reaction is $6.005* 10^(-6)$

Explanation:

We are given:

Initial moles of $NH_3=0.0150mol$

Initial moles of $O_2=0.0150mol$

Volume of the container = 1.00 L

Molarity of the solution = $\frac{\text{Number of moles}}{\text{Volume of container}}$

$[NH_3]_i=(0.0150)/(1.00)=0.0150M$

$[O_2]_i=(0.0150)/(1.00)=0.0150M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

Initial: 0.0150 0.0150

At eqllm: 0.0150-4x 0.0150-3x 2x 6x

The expression of $K_c$ for above equation follows:

$K_c=([N_2]^2[H_2O]^6)/([NH_3]^4[O_2]^3)$ .......(1)

We are given:

Equilibrium concentration of $N_2=1.96* 10^(-3)$

Equating the equilibrium concentrations of nitrogen, we get:

$2x=1.96* 10^(-3)\n\nx=0.98* 10^(-3)M$

Calculating the equilibrium concentrations:

Concentration of $NH_3=(0.0150-4x)=0.0150-4(0.00098)=0.01108M$

Concentration of $O_2=(0.0150-3x)=0.0150-3(0.00098)=0.01206M$

Concentration of $N_2=2x=2(0.00098)=0.00196M$

Concentration of $H_2O=6x=6(0.00098)=0.00588M$

Putting values in expression 1, we get:

$K_c=((0.00196)^2* (0.00588)^6)/((0.01108)^4* (0.01206)^3)\n\nK_c=6.005* 10^(-6)$

Hence, the value of $K_c$ for the reaction is $6.005* 10^(-6)$

Final answer:

To calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2, you need to determine the equilibrium concentrations of NH3 and O2. The given information includes the initial moles and concentration of NH3 and O2, as well as the equilibrium concentration of N2. Using the stoichiometry of the reaction and the given data, you can calculate the equilibrium concentrations and substitute them into the Kc expression to determine the numerical value of Kc.

Explanation:

The question asks to calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2. The reaction equation is 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) + 6 H2O(g). The given information is that 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00-L container, and the N2 concentration at equilibrium is 1.96 × 10−3 M. To solve for Kc, we need to calculate the equilibrium concentrations of NH3 and O2.

Using the stoichiometry of the reaction, we can determine that the equilibrium concentration of NH3 is (0.0150 - 2*1.96 × 10−3) M and the equilibrium concentration of O2 is (0.0150 - 3*1.96 × 10−3) M. Substituting these values into the equilibrium expression for Kc, we can calculate the value of Kc.

In this case, the equilibrium constant, Kc, can be calculated as [N2]^2 / ([NH3]^4 * [O2]^3). Substitute the given equilibrium concentration of N2 and the calculated equilibrium concentrations of NH3 and O2 into the Kc expression to determine the numerical value of Kc.

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It is a vocal music of Mindoro

Answers

Answer:

instrumental music is used in festivals, rituals, etc.

Explanation:

How do test with crash dummies, seat belts, and air bags illustrate newton's first law of motion??

Answers

Answer:

Newton’s law of inertia is illustrated in tests with crash dummies, seat belts, and airbags, wherein the object stays in motion unless there is an unbalanced force applied to it.

Inertia is the main reason why there are seatbelts and airbags in the car. In this case, when the seatbelt is trapped to the passenger, the passenger experiences the same state of motion as the car. If the car accelerates/decelerates, the passenger experiences it too. When the car experiences collision, an unbalance force is acted upon it. This causes the car to stop abruptly, and the passenger shares the same state of motion because of the seatbelt and the airbags that apply the unbalanced force to stop the passenger to go forward.

MassWhat is the percent composition by mass of
nitrogen in the compound N,H, (gram-formula
32 g/mol)?
(1) 13%
(3) 88%
(2) 44%
(4) 93%

Answers

Answer: 88%

Explanation:

The combustion of ethane (C2H6) produces carbon dioxide and steam. 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) How many moles of CO2 are produced when 5.90 mol of ethane is burned in an excess of oxygen?

Answers

Final answer:

The combustion of ethane yields carbon dioxide, and with 5.90 moles of ethane being reacted, it results in the production of 11.8 moles of CO2.

Explanation:

The question pertains to the concept of stoichiometry in chemistry, and the chemical reaction in question is a combustion reaction involving ethane (C2H6). From the balanced reaction, it is evident that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Therefore, if we have 5.90 moles of ethane reacting, it's a straightforward calculation to determine that this would yield twice that many moles of CO2. We simply multiply the moles of ethane by the stoichiometric ratio (4/2) to get the moles of CO2.

Example Calculation: 5.90 moles of ethane x (4 moles CO2 / 2 moles C2H6) = 11.8 moles CO2

So, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of CO2 are produced.

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Final answer:

In the combustion of ethane, for every mole of ethane burned, two moles of carbon dioxide are produced. Hence, when 5.90 moles of ethane are burned, 11.8 moles of carbon dioxide are produced.

Explanation:

The chemical reaction given, 2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g), states that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Thus, the mole-to-mole ratio of ethane to carbon dioxide is 2:4, or simplified, 1:2. So, for every mole of ethane burned, two moles of carbon dioxide are produced.

Given that 5.90 moles of ethane are burned, we can calculate the quantity of carbon dioxide produced by multiplying 5.90 moles by 2. Hence, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of carbon dioxide are produced.