The correct answer is option 1. Carbon dioxide is nonpolar because the shape of the molecule is symmetrical. It is a linear molecule where the oxygen atoms are symmetrical on each end. CO2 molecule do not have a region of unequal sharing.

7 0

3 years ago

A sample of gas contains 0.1900 mol of CO(g) and 0.1900 mol of NO(g) and occupies a volume of 22.0 L. The following reaction tak

worty [1.4K]

Answer:

V₂ = 16.5 L

Explanation:

To solve this problem we use Avogadro's law, which applies when temperature and pressure remain constant:

V₁/n₁ = V₂/n₂

In this case, V₁ is 22.0 L, n₁ is [mol CO + mol NO], V₂ is our unknown, and n₂ is [mol CO₂ + mol N₂].

n₁ = mol CO + mol NO = 0.1900 + 0.1900 = 0.3800 mol

We use the reaction to calculate n₂:

2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)

mol CO₂:

0.1900 mol CO * $\frac{2molCO_{2}}{2molCO} =$ 0.1900 mol CO₂

mol N₂:

0.1900 mol NO * $\frac{1molN_{2}}{2molNO} =$ 0.095 mol N₂

n₂ = mol CO₂ + mol N₂ = 0.1900 + 0.095 = 0.2850 mol

Calculating V₂:

22.0 L / 0.3800 mol = V₂ / 0.2850 mol

V₂ = 16.5 L

3 0

4 years ago

Which property to put in which

NeTakaya

In the Proton category: symbol p, and +1 charge
In the Neutron category: symbol n, 0 charge
In the Electron category: symbol e, -1 charge

6 0

3 years ago

In a generic chemical reaction involving reactants A and B and products C and D, aA + bB → cC + dD, the standard enthalpy ΔH∘rxn

velikii [3]

Answer:

A. ΔH∘rxn = 21.9 KJ/mol

B. ΔH∘rxn = 103 KJ/mol

C. C2H5OH + 3O2 → 2CO2 + 3H2O

Explanation:

The standard reaction equation is given as:

aA + bB → cC + dD

Its standard enthalpy is given as:

ΔH∘rxn = cΔH∘f(C) + dΔH∘f(D) − aΔH∘f(A) − bΔH∘f(B)

Reaction given to us is:

H2O(l) + CCl4(l) → COCl2(g) + 2HCl(g)

So, its standard enthalpy will be:

ΔH∘rxn = (1)ΔH∘f(CoCl2 (g)) + (2)ΔH∘f(HCl(g)) − (1)ΔH∘f(H2O(l)) − (1)ΔH∘f(CCl4(l))

using the values from table:

ΔH∘rxn = - 218.8 KJ/mol + (2)(- 92.3 KJ/mol) - (- 285.8 KJ/mol) - (- 139.5 KJ/mol)

ΔH∘rxn = 21.9 KJ/mol

Reaction given to us is:

2A + B ⇌ 2C + 2D

So, its standard enthalpy will be:

ΔH∘rxn = (2)ΔH∘f(C) + (2)ΔH∘f(D) − (2)ΔH∘f(A) − (1)ΔH∘f(B)

using the values from table:

ΔH∘rxn = (2)181 KJ/mol + (2)(- 523 KJ/mol) - (2)(- 225 KJ/mol) - (- 337 KJ/mol)

ΔH∘rxn = 103 KJ/mol

Balanced equation for combustion of ethanol is:

C2H5OH + 3O2 → 2CO2 + 3H2O

3 0

3 years ago