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3 years ago

7

How are ionic compound formed? when atoms share electrons to complete eight valence electrons when atoms lose electrons and form

positive ions when electrons are transferred from one atom to another forming positive and negative ions that atract each other submit rewatch?

1 answer:

Pepsi [2]3 years ago

3 0

Ionic Crystal. Ionic hydrogen.postive electron transfer to atoms

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Which of the questions below is scientific? A. What is my blood pressure right now? B. How does regular exercise affect blood pr

melamori03 [73]

Answer:

B

Explanation:

the reason as to why B is the answer is because, the question is one that you will have to study and find more information about, while the others are simple and do not require you to go into such detail and research.

7 0

3 years ago

The standard enthalpy change for the reaction of SO3(g) with H2O(l) to yield H2SO4(aq) is ΔH∘ = -227.8 kJ . Use the following in

olga nikolaevna [1]

Answer:

-909.3KJ/mole

Explanation:

The heat of reaction is accessible from the heat of formation of reactants and products using the formula below:

ΔH = Σ ΔHf products - Σ ΔHf reactants

Before we proceed, it is important to know that the enthalpy of formation of element is zero ,be it a single element or a molecule of an element.

From the reaction for the formation of sulphuric acid, we know we need to know the heat of formation of sulphur (vi) oxide and water. The examiner is quite generous and have us for water already.

Now we need to calculate for sulphur (vi) oxide. This is calculated as follows:

We first calculate for sulphur(iv)oxide. This can be obtained from the reaction between sulphur and oxygen. The calculation goes thus:

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole suphur(iv) oxide × x] - [ (1 mole of elemental sulphur × 0) + (1 mole of elemental oxygen × 0]

We were already told this is equal to -296.8KJ. Hence the heat of formation of sulphur(iv) oxide is -296.8KJ.

We then proceed to the second stage.

Now, here we have 1 mole sulphur (iv) oxide reacting with 0.5 mole oxygen molecule.

We go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphur (vi) oxide × y] - [ (1 mole of sulphur (iv) oxide × -296.8) + (0.5 mole of oxygen × 0)].

We already know that the ΔH here equals -98.9KJ.

Hence, -98.9 = y + 296.8

y = -296.8KJ - 98.9KJ = -395.7KJ

We now proceed to the final part of the calculation which ironically comes first in the series of sentences.

Now, we want to calculate the standard heat of formation for sulphuric acid. From the reaction, we can see that one mole of sulphur (vi) oxide, reacted with one mole of water to yield one mole of sulphuric acid.

Mathematically, we go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphuric acid × z] - [( 1 mole of sulphur vi oxide × -395.7) + ( 1 mole of water × -285.8)].

Now, we know that the ΔH for this particular reaction is -227.8KJ

We then proceed to to open the bracket.

-227.8 = z - (-395.7 - 285.8)

-227.8 = z - ( -681.5)

-227.8 = z + 681.5

z = -227.8-681.5 = -909.3KJ

Hence, ΔH∘f for sulphuric acid is -909.3KJ/mol

6 0

3 years ago

How many moles are in 25 grams of HF

anzhelika [568]

$M_{HF}=20\frac{g}{mol}\\
m=25g\\\\
n=\frac{m}{M}=\frac{25g}{20\frac{g}{mol}}=1,25mol$

4 0

4 years ago

In the redox persulfate-iodide experiment, a 100 mL reaction mixture is prepared for one of the runs as follows. 0.200 M KI 20 m

grigory [225]

Answer:

a) The number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.

b) The number of moles of S₂O₈2- that would react to produce the blue color will be 0.00005 M.

c) rate = 2.5 x 10-6M/L/s

Explanation:

a)

The reactions taking place in this experiment are represented by the ionic equations,

S₂O₈ 2- + 2I- ----------> 2SO₄2- + I₂ --------------------(1)

2 S₂O₃2- + I₂ -----------> S₄O₆ +2 I- -------------------(2)

The persulphate ions react with the iodide ions to produce free iodine which is in turn reduced by the thiosulphate ions to produce iodide ions again. This reaction proceeds till all the thiosulphate ions are used up. Therefore the rate of the reaction will be the rate at which iodine is formed and used up.

When there is free iodine the reaction mixture, the solution gives a dark blue coloration. This happens when all the thiosulphate ions are used up.

The volume of sodium thiosulphate ( Na₂S₂O3) solution added to the reaction vessel = 10ml

Molarity of sodium thiosulphate ( Na₂S₂O3) solution = 0.01M

Number of moles of ( Na2S2O3) = 0.01ml x 10M /1000ml = 0.0001M (molarity x volume in L)

The number of moles of thiosulphate reacted will be equal to the number of moles taken since the reaction proceeds till all the thiosulphate is consumed.

Hence, the number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.

b)

To calculate S₂O₈ 2-

The permanent blue color is produced once all the thiosulphate ions are used up and persulphate reacts with iodide ions to produce iodine, so, the number of moles of persulphate ions will be equal to the number of moles of iodine formed

According to the stoichiometry of equation 1.

1 mole of S₂O₈ 2-produces 1 mole of iodine.

According to the stoichiometry of equation 2,

1mole of iodine produced consumes 2 moles of S₂O₃2-

The number of moles of S₂O₃2- taken = 0.0001M.

2 moles of S₂O₃2- is equivalent to 1 mole I₂

therefore

0.0001 mole of S₂O₃2- = 0.0001/2 = 0.00005M of I₂

Since stoichimetrically,

1 mole of S₂O₈2- is equivalent to 1 mole I2, the number of moles of S₂O₈2- that would react to produce the blue color will be 0.00005 M.

c)

The initial reaction rate is given by

rate =change in concentration of persulphate ion [S₂O₈2-] / time

rate = change in Concentraion of I₂ / t

since initial concentration of I₂ = 0.

rate = Concentraion of I₂/ t

The concentration of I₂ = number of moles of iodine / total volume of solution in L

= 0.00005M/ 0.1L = 0.0005M/L (Volume of the solution = 100ml = 0.1L)

rate = Concentraion of I₂ / t

= 0.0005 /200s

rate = 2.5 x 10-6M/L/s

4 0

3 years ago

I want to know the steps.

Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(Δ $H_{f}$ of $C_{6}H_{12} O_{6}$ ) =-1275.0

enthalpy of combustion of oxygen(Δ $H_{f}$ of $O_{2}$ ) = zero

enthalpy of combustion of carbon dioxide(Δ $H_{f}$ of $CO_{2}$ ) = -393.5

enthalpy of combustion of water(Δ $H_{f}$ of $H_{2} O$ ) = -285.8

To solve :

standard enthalpy of combustion

We know;

Δ $H_{f}$ = ∈Δ $H_{f}$ (products) - ∈Δ $H_{f}$ (reactants)

$C_{6}H_{12} O_{6}$ (s) +6 $O_{2}$ (g) → 6 $CO_{2}$ (g)+ 6 $H_{2} O$ (l)

Δ $H_{f}$ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

Δ $H_{f}$ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

Δ $H_{f}$ = 6 (-393.5) + 6(-285.8) - 0 + 1275

Δ $H_{f}$ = -2361 - 1714 - 0 + 1275

Δ $H_{f}$ =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0

3 years ago