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4 years ago

Which statement is true about crystal lattice energy

Chemistry

2 answers:

expeople1 [14]4 years ago

6 0

It increases as the size of the ions increases

Arturiano [62]4 years ago

3 0

Hey there!:

1) The additional stability that accompanies the formation of the networkCrystalline is measured as network enthalpy.
2) The reticular energy is the energy released when the solid Crystal isform from separate ions in the gaseous state. Always exothermic.
3) The enthalpy of the network depends directly on the size of the loads and conversely in the distance between the ions .

hope this helps!

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Which particle may be gained, lost, or shared by an atom when it forms a chemical bond?

andrew-mc [135]

Electrons because the amount of valence electrons determines the bonds it can form and often times during a chemical bond or the forming of a compound an element will lose some of its electrons.

4 0

3 years ago

One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable.a. 39/19K an

marishachu [46]

Answer:

a) 39/19 K : stable nuclide, 40/19 K : radioactive nuclide.

b) 209B: stable nuclide, 208Bi : radioactive nuclide

c) nickel-58 : stable nuclide, nickel-65 : radioactive nuclide.

Explanation:

As per the rule, nuclides having odd number of neutrons are generally not stable and therefore, are radioactive.

Mass number (A) = Atomic number (Z) + No. of neutrons (N)

Or, N = A - Z

39/19 K and 40/19 K

Calculate no. of neutrons in 39/19 K as follows:

atomic no. = 19, mass no. 39

N = 39 - 19

= 20 (even no.)

Calculate no. of neutrons in 40/19 K as follows:

atomic no. = 19, mass no. 40

N = 40 - 19

= 21 (odd no.)

Therefore, 39/19 K is a stable nuclide and 40/19 K is a radioactive

nuclide.

209Bi and 208Bi

Calculate no. of neutrons in 209Bi as follows:

atomic no. = 83, mass no. 209

N = 209 - 83

= 126 (even no.)

Calculate no. of neutrons in 208Bi as follows:

atomic no. = 83, mass no. 208

N = 208 - 83

= 125 (odd no.)

Therefore, 209Bi is a stable nuclide and 208Bi is a radioactive nuclide.

nickel-58 and nickel-65

Calculate no. of neutrons in nickel-58 as follows:

atomic no. = 28, mass no. 58

N = 58 - 28

= 30 (even no.)

Calculate no. of neutrons in nickel as follows:

atomic no. = 28, mass no. 65

N = 65 - 28

= 37 (odd no.)

Therefore,nickel-58 is a stable nuclide and nickel-65 is a radioactive nuclide.

5 0

3 years ago

Why does blowing carbon dioxide gas into aqueous barium hydroxide reduce?

Oksana_A [137]

If one chooses to blow carbon dioxide gas into aqueous barium hydroxide it will reduce. This happens because the ions become removed from the solution and in turn this decreases the conductivity of it.

5 0

4 years ago

Read 2 more answers

1. The electron arrangement ions X and Y2 are 2, 8, and 2, 8, 8 respectively.

svetoff [14.1K]

The given question is incomplete. the complete question is:

The electron arrangement ions $X^{3+}$ and $Y^{2-}$ are 2, 8, and 2, 8, 8 respectively.

Write the electronic arrangement of the elements X and Y.

Write the formula of the compound that would be formed between X and Y.

Answer: 1. The electronic arrangement for X and Y are 2,8, 3 and 2,8,5 respectively.

The formula of the compound formed is $X_2Y_3$

Explanation:

Electronic configuration is the arangement of electrons in an atom in order of increasing energies.

Cations are formed when electrons are lost by atoms and anions are formed when electrons are gained by atoms.

$X^{3+}:2,8$

$X:2,8,3$

$Y^{3-}:2,8,8$

$Y:2,8,5$

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

Here element X is having an oxidation state of +3 called as $X^{3+}$ cation and $Y^{2-}$ is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $X_2Y_3$

3 0

3 years ago

The starting substanceses of a chemical reaction are refered to as

Debora [2.8K]

The starting substances in a chemical reaction are called reactants - they are written on the left side of a chemical equation.

6 0

4 years ago