What is a substance?

NItrogen in air reacts at high temperature to form NO2 according to the reaction:

Rina8888 [55]

Answer:

The correct answer is option E.

Explanation:

Structures for the reactants and products are given in an aimage ;

Number of double bonds in oxygen gas molecule = 1

Number of double bonds in nitro dioxide gas molecule = 1

Number of single bond in in nitro dioxide gas molecule = 1

Number of triple bonds in nitrogen gas molecule = 1

$N_2+2O_2\rightarrow 2NO_2,\Delta H=?$

$\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]$

$\Delta H_{f,NO_2}=33.18 kJ/mol$

$\Delta H_{f,N_2}=0$ (pure element)

$\Delta H_{f,O_2}=0$ (pure element )

$\Delta H=2 mol\times 33.18 kJ/mol=66.36kJ=15.86 kcal$

The enthalpy of the given reaction is 15.86 kcal.

6 0

3 years ago

PLEASE HELP I WILL MARK BRAINLIEST!!! Reactions In Our World Lab Report

Cerrena [4.2K]

Answer:

ok

Explanation:

ok

6 0

2 years ago

Consider the titration of sulfuric acid with sodium hydroxide. What volume (mL) of a 2.658M NaOH solution is required to fully t

KatRina [158]

The volume of base required to completely neutralize the acid is 3.2 mL of NaOH.

The equation of the reaction is;

2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)

From the question;

Concentration of acid CA = 0.426M

Concentration of base CB = 2.658M

Volume of acid VA = 10.00mL

Volume of base VB = ?

Number of moles of acid NA = 1

Number of moles of base NB = 2

Using the relation;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

VB = CAVANB/CBNA

VB = 0.426M × 10.00mL × 2/ 2.658M × 1

VB = 3.2 mL

Learn more: brainly.com/question/6111443

8 0

2 years ago

Which of the following is NOT a characteristic of stars? A) Size B) temperature C) texture D) color

Lynna [10]

Answer:

C

Explanation:

Characteristics used to classify stars include color, temperature, size, composition, and brightness.

7 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago