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Dovator [93]
2 years ago
14

A gaseous compound is 30.4% nitrogen and 69.6% oxygen. a 6.06-gram sample of gas occupies a volume of 1.00 liter and exerts a pr

essure of 1.26 atmospheres at – 40.0°c. what is its molecular formula?
Chemistry
1 answer:
aksik [14]2 years ago
8 0

A gaseous compound is 30.4% nitrogen and 69.6% oxygen by mass. A 5.25-g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 1.26 atm at -4.0°C. Which of the following is its molecular formula?  

1) NO2  

2) N3O6  

3) N2O5  

4) N2O4  

5) NO

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Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2
Svet_ta [14]

Answer:

-1815.4 kJ/mol

Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol

4 0
2 years ago
2 attempts left select the single best answer. The radii of the lithium and magnesium ions are 76 pm and 72 pm, respectively. Wh
Lina20 [59]

The ionic character of any compound depend on the lattice energy as well as the electronegativity of element present in that compound.

More would be the lattice energy more would be ionic nature of that compound.

The lattice energy of any compound is inversely proportional to the ionic radii cation and anion.

In given case the ionic radii of oxide in both oxides would be equal therefore the lattice energy only depend on the ionic radii of cation.

Lattice energy (U)     =   \frac{1}{Ionic radii}

As the radii of Magnesium less then radii of lithium therefore lattice energy of Magnesium oxide would be more than lithium oxide.

Hence, MgO would be more ionic in nature than Li_{2}O

8 0
2 years ago
You are about to watch a video about a scientist who studies jellies like the one shown in this image. One thing she investigate
Dvinal [7]
Predators decreased, food source increased, migration patterns, natural disaster or threat in normal habitat.
5 0
2 years ago
Which statement is TRUE about polar covalent bond?
Anvisha [2.4K]
I think the correct answer among the choices listed above is option D. Polar covalent bond is seen in molecules where <span>electrons are not shared equally between atoms. This makes the molecule to have a partially positive and partially negative sides.</span>
3 0
2 years ago
What is the value of ΔGo in kJ at 25 oC for the reaction between the pair: Mn(s) and Ag+(aq) to give Ag(s) and Mn2+(aq) Use the
Leni [432]

Answer:  -3.8\times 10^{5}J

Explanation:

Mn+2Ag^{+}\rightarrow Mn^{2+}+2Ag

Here Mn undergoes oxidation by loss of electrons, thus act as anode. silver undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0_{[Mn^{2+}/Mn]}= -1.18V

E^0_{[Ag^{2+}/Ag]}=+0.80V

E^0=E^0_{[Ag^{+}/Ag]}- E^0_{[Mn^{2+}/Mn]}

E^0=+0.80- (-1.18V)=1.98V

The standard emf of a cell is related to Gibbs free energy by following relation:

\Delta G^0=-nFE^0

\Delta G^0 = gibbs free energy

n= no of electrons gained or lost  = 2

F= faraday's constant

E^0 = standard emf  = 1.98V

\Delta G^0=-2\times 96500\times (1.98)=-3.8\times 10^{5}J

Thus the value of \Delta G^0 is -3.8\times 10^{5}J

8 0
2 years ago
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