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3 years ago

State how you can tell from a dot and cross diagram that the particles in a compound are held together by ionic bonds

Chemistry

1 answer:

Darina [25.2K]3 years ago

8 0

Answer:

In diagram you have to show the electrons(dots) near to the more electronegative element by this you can show that this is an ionic bond.

Explanation:

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For the reaction, Cl2 + 2KBr --> 2KCl + Br2, how many moles of potassium chloride, KCl, are produced from 102g of potassium b

Sedaia [141]

Solution :

From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .

Moles of KBr in 102 g of potassium bromide.

n = 102/119.002

n = 0.86 mole.

So, number of miles of KCl produced are also 0.86 mole.

Mass of KCl produced :

$m = 0.86 \times Molar \ mass \ of \ KCl\\\\m = 0.86 \times 74.5 \ gram \\\\m = 64.07\ gram$

Hence, this is the required solution.

5 0

2 years ago

Given the equation: HCl + Na2SO4 → NaCl + H2SO4, if you start with 8 moles of hydrochloric acid, how many grams of sulfuric acid

Simora [160]

Answer:

392g sulfuric acid are produced

Explanation:

Based on the balanced equation:

2HCl + Na2SO4 → 2NaCl + H2SO4

2 moles of HCl produce 1 mole of sulfuric acid

To solve the problem we need to find the moles of sulfuric acid produced based on the chemical equation. Then, using its molar mass -Molar mass H2SO4 = 98g/mol- we can find the mass of sulfuric acid produced:

Moles sulfuric acid:

8mol HCl * (1mol H2SO4 / 2mol HCl) = 4 mol H2SO4

Mass sulfuric acid:

4mol H2SO4 * (98g / mol) =

392g sulfuric acid are produced

4 0

3 years ago

It takes to break an iodine-iodine single bond. Calculate the maximum wavelength of light for which an iodine-iodine single bond

Zolol [24]

The given question is incomplete. The complete question is :

It takes 151 kJ/mol to break an iodine-iodine single bond. Calculate the maximum wavelength of light for which an iodine-iodine single bond could be broken by absorbing a single photon. Be sure your answer has the correct number of significant digits.

Answer: 793 nm

Explanation:

The relation between energy and wavelength of light is given by Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

E = energy of the light = 151 kJ= 151000 J (1kJ=1000J)

N= moles = 1 = $6.023\times 10^{23}$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light = ?

Putting in the values:

$151000J=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{\lambda}$

${\lambda}=7.93\times 10^{-7}m=793nm$ $1m=10^{-9}nm$

Thus the maximum wavelength of light for which an iodine-iodine single bond could be broken by absorbing a single photon is 793 nm

3 0

2 years ago

Consider the following system at equilibrium:A(aq)+B(aq) <---> 2C(aq)Classify each of the following actions by whether it

velikii [3]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On addition of reactant at equilibrium shifts the equilibrium in forward direction.
On addition of product at equilibrium shifts the equilibrium in backward direction.
On removal of reactant at equilibrium shifts the equilibrium in backward direction.
On removal of product at equilibrium shifts the equilibrium in forward direction.

$A(aq)+B(aq)\rightleftharpoons 2C(aq)$

Reactants = A , B

Product = C

1. Increase A

On increasing the amount of A at equilibrium will shift the equilibrium in forward or rightward direction.

2. Increase B

On increasing the amount of B at equilibrium will shift the equilibrium in forward or rightward direction.

3. Increase C

On increasing the amount of C at equilibrium will shift the equilibrium in backward or leftward direction.

4. Decease A

On decreasing the amount of A at equilibrium will shift the equilibrium in backward or leftward direction.

5. Decease B

On decreasing the amount of B at equilibrium will shift the equilibrium in backward or leftward direction.

6. Decease C

On decreasing the amount of C at equilibrium will shift the equilibrium in forward or rightward direction.

7. Double A and Halve B

Equilibrium constant of the reaction = K

$K=\frac{[C]^2}{[A][B]}$

On doubling A and halving B, equilibrium constant of the reaction = K'

$K'=\frac{[C]^2}{[2A][\frac{B}{2}]}=\frac{[C]^2}{[A][B]}$

The value of equilibrium constant K' is equal to K, which means that equilibrium will not shift in any direction.

8. Double both B and C

Equilibrium constant of the reaction = K

$K=\frac{[C]^2}{[A][B]}$

On doubling B and C, equilibrium constant of the reaction = K'

$K'=\frac{[2C]^2}{[A][2B]}=\frac{4[C]^2}{[A][2B]}=\frac{2[C]^2}{[A][B]}$

K' = 2 K

The value of equilibrium constant K' is double the K, which means that product is increasing which means that equilibrium will shift in backward or leftward direction.

5 0

3 years ago

It takes 60 mL of 0.20 M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H2CO3) for the following chemical reac

irina1246 [14]

If It takes 60mL of 0.20M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid ( $H_2CO_3$ ), then the concentration of the carbonic acid is 0.24M