Steel is an alloy of one or more elements combined with which metal?

The atom of chlorine has several valence electrons in its

Ivan

The atom of chlorine has several valence electrons in its valence electrons are in the third shell.

8 0

3 years ago

1. Identify the atom in that has the greater ionization energy. A. Calcium Ca B. Barium Ba

Alisiya [41]

Answer:

Calcium

Explanation:

Calcium-6,1132

Barium-5,2117

6 0

3 years ago

Read 2 more answers

Hunter is copying an angle. His work so far follows. Explain the importance of his next step. Which is drawing A-line through A

forsale [732]

Answer:

its c ma man .

This is the other ray that will make up the angle ∠AYZ and will complete the constructioExplanation:

if your taking the quick check these were my awnsers

1.c

2.b

3.d

4.c

5 .A

for the copying an angle quick check only

6 0

3 years ago

Which of the following equilibrium expressions corresponds to a heterogeneous equilibrium?(a) Kc =[NH4+][OH−][NH3](b) Kc =[H+][C

Alex777 [14]

Answer:

c. $Kc = \frac{[Ag(NH_{3})_2^{+}][Cl^-]}{[NH_3]^2}}$

d. $Kc = [Ba^{2+}][F^{-}]^2$

Explanation:

An heterogeneous equilibrium is defined as a system whose reactants, products, or both are in more than one phase.

Also, you must know in an equilibrium constant you don't take solids or pure liquids. Thus:

a. $Kc = \frac{[NH_{4}^+][OH^-]}{[NH_3]}}$

The reaction must be: NH₃(aq) + H₂O(l) ⇆ NH₄⁺(aq) + OH⁻(aq) All reactants are in the same phase.

b. $Kc = \frac{[H^+][C_2H_3O_{2}^-]}{[HC_2H_3O_2]}}$

The reaction must be: HC₂H₃O₂(aq) ⇄ H⁺(aq) + C₂H₃O₂⁻(aq) All reactants are in the same phase.

c. $Kc = \frac{[Ag(NH_{3})_2^{+}][Cl^-]}{[NH_3]^2}}$

The reaction must be: 2NH₃(aq) + AgCl(s) ⇄ Ag(NH₃)₂⁺(aq) + Cl⁻(aq)

Reactants aren´t in the same phase. Heterogeneous equilibrium

d. $Kc = [Ba^{2+}][F^{-}]^2$

The reaction must be: BaF₂(s) ⇆ Ba²⁺(aq) + 2F⁻(aq)

Reactants aren´t in the same phase. Heterogeneous equilibrium

I hope it helps!

6 0

3 years ago

Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC

Anvisha [2.4K]

Answer: The $\Delta G$ for the reaction is 55.328 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

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

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.100atm\\p_{CCl_4}=0.180atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.100\times 0.180}\\\\K_p=30.0125$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

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

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

$K_p$ = equilibrium constant in terms of partial pressure = 30.0125

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(30.0125))\\\\\Delta G=55327.74J/mol=55.328kJ/mol$

Hence, the $\Delta G$ for the reaction is 55.328 kJ/mol

5 0

3 years ago