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Liula [17]
2 years ago
12

2 attempts left select the single best answer. The radii of the lithium and magnesium ions are 76 pm and 72 pm, respectively. Wh

ich compound has stronger ionic attraction, lithium oxide or magnesium oxide? Lithium oxide magnesium oxide both are equal.
Chemistry
1 answer:
Lina20 [59]2 years ago
8 0

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

You might be interested in
Instrumental methods need only a microscopic sample to return an accurate result. Why is this so?
uranmaximum [27]
Instrumental methods of analysis rely on machines.The visualization of single molecules, single biological cells, biological tissues and nanomaterials is very important and attractive approach in analytical science.
There are several different types of instrumental analysis. Some are suitable for detecting and identifying elements, while others are better suited to compounds. In general, instrumental methods of analysis are:
-Fast
-Accurate (they reliably identify elements and compounds)
-Sensitive (they can detect very small amounts of a substance in a small amount of sample)
3 0
2 years ago
A 0.3870-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.7191Â g of co
olchik [2.2K]
The chemical formula for the compound can be written as,

    CxHyOz 

where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
   
    CxHyOz + O2 --> CO2 + H2O 

number of moles of C:
     (0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
 This signifies that 0.0163 mole of C and the mass of carbon in the compound,
        (0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C

number of moles H:
      (0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O

This signifies that there are 0.01635 atoms of H in the compound.
      mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H

Mass of oxygen in the compound,
   0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g

Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O

The formula of the compound is,
      C0.0163H0.01635O0.0109

Dividing the numbers by the least number,
    C3/2H3/2O

The empirical formula of the compound is therefore,
    <em>  C₃H₃O₂</em>
5 0
2 years ago
Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3
ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)×\frac{1molZn}{65,38g}×\frac{1molH_{2}(g)}{1molZn} = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = <em>190,1L</em>

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL×\frac{101,325}{1atmL} = <em>19,26 kJ</em>

I hope it helps!

5 0
2 years ago
A vessel contains Ar(g) at a high pressure. Which of the following statements best helps to explain why the measured pressure is
Luda [366]

Answer:

4. The combined volume of the Ar atoms is too large to be negligible compared with the total volume of the container.

Explanation:

Deviations from ideality are due to intermolecular forces and to the nonzero volume of the molecules themselves. At infinite volume, the volume of the molecules themselves is negligible compared with the infinite volume the gas occupies.

However, the volume occupied by the gas molecules must be taken into account. Each <u>molecule does occupy a finite, although small, intrinsic volume.</u>

The non-zero volume of the molecules implies that instead of moving in a given volume V they are limited to doing so in a smaller volume. Thus, the molecules will be closer to each other and repulsive forces will dominate, resulting in greater pressure than the one calculated with the ideal gas law, that means, without considering the volume occupied by the molecules.

5 0
2 years ago
The decomposition of nitramide, O 2 NNH 2 , O2NNH2, in water has the chemical equation and rate law O 2 NNH 2 ( aq ) ⟶ N 2 O ( g
valkas [14]

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1  O ₂N N H ⁻ + H ⁺ ( f a s t  e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ]  / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

r a t e = k ₂ k ₁/ k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]

Therefore, the overall rate constant is

k = k₂k₁/k-1

5 0
1 year ago
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