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3 years ago

Explain how atoms(ions) are held together in an ionic bond. Give an example of an ionic compound

2 answers:

7 0

Ionic bond is described as the chemical bond between two oppositely charged ions. In ionic bond the metal loses electron forming a positively charged cation and a non metal which accept the cation to form a negatively charged anion. In ionic bond the atoms are held together by electrostatic force of attraction. In ionic bond the anions and cations are present in the ratio where the total charge of the compound becomes zero. For example, Let us consider NaCl compound. An atom of the sodium has three electrons in its valency shell and the electrons are removed from the outer most shell by applying the energy of 5.14 electron volts. The chlorine atom lacks an electron to attain stable electronic configuration and it accepts the electron from the sodium by releasing 3.62 electron volts of energy which means that it takes only 1.52 electron volts of energy to donate an electron to chlorine when both the atoms are far apart. When these electrons are brought together their electric potential becomes more negative. This means that if the neutral sodium and chlorine atoms are found themselves closer it would be energetically favourable to transfer electrons from sodium to chlorine thus resulting in the formation of the ionic bond.

fgiga [73]3 years ago

5 0

The ions in an ionic bond are held together by electrostatic attraction between the anions and cations in the compound.

For example, sodium chloride [NaCl] (more commonly known as salt) has positively charged sodium [Na+] ions, or cations, and negatively charged chlorine [Cl-] ions, or anions, holding the compound together.

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4 years ago

Read 2 more answers

The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: CO(g) + H2O(g)

taurus [48]

<u>Answer:</u> The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

<u>Explanation:</u>

For the given chemical reaction:

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}$ ........(1)

We are given:

$[CO]_{eq}=[H_2O]_{eq}=[H_2]_{eq}=0.10M$

$[CO_2]_{eq}=0.40M$

Putting values in above equation, we get:

$K_c=\frac{0.40\times 0.10}{010\times 0.10}\\\\K_c=4$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of hydrogen gas = 0.30 mol

Volume of solution = 2.0 L

Putting values in above equation, we get:

$\text{Molarity of }H_2=\frac{0.30mol}{2L}=0.15M$

When hydrogen gas is added, the concentration of product gets increased. But, by Le-Chatelier's principle, the equilibrium will shift in the direction where concentration of product must decrease, which is in the backward direction.

Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M

Now, the equilibrium is shifting to the reactant side. The equation follows:

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial: 0.1 0.1 0.4 0.1

At eqllm: 0.1+x 0.1+x 0.4-x 0.25-x

Putting values in expression 1, we get:

$4=\frac{(0.25-x)(0.4-x)}{(0.1+x)(0.1+x)}\\\\3x^2+1.45x-0.06=0\\\\x=0.038,-0.522$

Neglecting the negative value of 'x'

Calculating the concentrations of the species:

Concentration of carbon dioxide = (0.4 - x) = (0.4 - 0.038) = 0.362 M

Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M

Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Hence, the concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

8 0

4 years ago

g 32.53 g of a solid is heated to 100.oC and added to 50.0 g of water in a coffee cup calorimeter and the contents are allowed t

Radda [10]

Answer:

0.886 J/g.°C

Explanation:

Step 1: Calculate the heat absorbed by the water

We will use the following expression

Q = c × m × ΔT

where,

Q: heat

c: specific heat capacity

m: mass

ΔT: change in the temperature

Q(water) = c(water) × m(water) × ΔT(water)

Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J

According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.

Q(water) + Q(solid) = 0

Q(solid) = -Q(water) = -1.89 × 10³ J

Step 2: Calculate the specific heat capacity of the solid

We will use the following expression.

Q(solid) = c(solid) × m(solid) × ΔT(solid)

c(solid) = Q(solid) / m(solid) × ΔT(solid)

c(solid) = (-1.89 × 10³ J) / 32.53 g × (34.4 °C - 100. °C) = 0.886 J/g.°C

8 0

3 years ago