Methane to acetylene??

Based on the chart, determine the identity of a substance that increases in temperature by 11.4oC when 1250J of energy are added

Firdavs [7]

Answer:

LIthium vegetable oil

7 0

3 years ago

6.Which set of coefficients correctly balances the following chemical equation?___ Mg(OH)2 + ___ HCl Imported Asset ___ MgCl2 +

scoundrel [369]

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

B. 1, 2, 1, 2

6 0

3 years ago

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Molecular chlorine and molecular fluorine combine to form a gaseous product. Under the same conditions of temperature and pressu

uranmaximum [27]

Answer:

Cl F₃

Explanation:

1) Reactants:

The reactants are:

Molecular chlorine: this is a gas diatomic molecule, i.e. Cl₂ (g)

Molecular fluorine: this is also a gas diatomic molecule: F₂ (g)

2) Stoichiometric coefficients:

One volume of Cl₂ react with three volumes of F₂ means that the reaction is represented with coefficients 1 for Cl₂ and 3 for F₂. So, the reactant side of the chemical equation is:

Cl₂ (g) + 3F₂ (g) →

3) Product:

It is said that the reaction yields two volumes of a gaseous product; then, a mass balance indicates that the two volumes must contain 2 parts of Cl and 6 parts of F. So, one volume must contain 1 part of Cl and 3 parts of F. That is easy to see in the complete chemical equation:

Cl₂ (g) + 3F₂ (g) → 2Cl F₃ (g)

As you see, that last equation si balanced: 2 atoms of Cl and 6 atoms of F on each side, and you conclude that the formula of the product is ClF₃.

8 0

2 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

Learn more:

6 0

2 years ago

Balance the equation 2Fe + 3Cl2 → 2FeCl3

Tresset [83]

Answer:

6 Cl0 + 6 e- → 6 Cl-I (reduction)

2 Fe0 - 6 e- → 2 FeIII (oxidation)

4 0

2 years ago

Methane to acetylene?? ​

Methane to acetylene??