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3 years ago

PLEASE I NEED THIS ANSWERED ASAP

Chemistry

1 answer:

kherson [118]3 years ago

7 0

The answer would be:
A. Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.

In this question, there are two half-reaction equations. To merge them up, you need to add the reactant with the reactant, then the product with the product. If there is a molecule on both side, you can cancel them. The full reaction would be:

C+ 1/2 O2 + CO + 1/2O2 ==>CO+ CO2 -----> remove CO from both side
C+ O2 ==>CO2

You might be interested in

Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper

KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

6 0

3 years ago

PLS HELP ASAP!!!! Describe metallic bonds, include an explanation of their valence electrons. How this related to the properties

Aloiza [94]

If you look it up it will give you plenty of information. This is what I found:

The valence electrons of metals move freely in this way because metals have relatively low electronegativity, or attraction to electrons. The positive metal ions form a lattice-like structure held together by all the metallic bonds. ... When nonmetals bond together, the atoms share valence electrons and do not become ions

https://www.ck12.org/c/physical-science/metallic-bond/lesson/Metallic-Bonding-MS-PS/

5 0

3 years ago

Which of these is NOT a colloid?

SVETLANKA909090 [29]

Answer:C

Explanation:

Mountain Dew soda is not a colloid

8 0

3 years ago

How many atoms are in 3.690 moles of oxygen

sweet [91]

1 mole of any substance contains Avogadro's number.

So, 1 mole of O2= 6.023x10^23 molecules

3 mole of O2= 6.023x10^23x3 molecules

= 1.8069x10^24 molecules

Each molecule of Oxygen has 2 atoms.

therefore,

1.8069x10^24 molecules= 1.8069x10^24 x 2 atoms

= 3.6138x10^24 atoms.

4 0

3 years ago

A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t

soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0

3 years ago