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vladimir2022 [97]
2 years ago
6

Gases react explosively if the heat released when the reaction begins is sufficient to cause more reaction, which leads to a rap

id expansion of the gases. Use bond energies to calculate ΔH°of each of the following reactions, and predict which occurs explosively:
(b) H₂(g) + I₂(g) → 2HI(g)
Chemistry
1 answer:
iogann1982 [59]2 years ago
3 0

∆H° of the following reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.

<h3>What is Bond Enthalpy? </h3>

The minimum amount of energy which is required to braak down or form the bonds in chemical reaction is known as bond enthalpy.

It can be calculated as:

∆Hrxn = sum of ∆H bond broken - sum. of ∆H of bond formed.

In order to Calculate ∆Hrxn for the given equation we have:

Bond energies in kJ/mol

  • H—H = 436
  • H—I = 295
  • I—I = 151

Now, the given reaction is

H₂(g) + I₂(g) → 2HI(g)

Here, 1 mol of H₂ and 1 mole of I₂ breaks to form 2 moles of HI.

Therefore,

We know that,

∆Hrxn = B. E(H—H) + B. E(I—I) - 2B. E(H—I)

= 436 + 151 - 2× 295

= 436+ 151 - 590

∆Hrxn = -3kJ/mol.

Thus, from the above conclusion we can say that ∆Hrxn of the reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.

learn more about Bond energy:

brainly.com/question/26964179

#SPJ4

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Komok [63]

Answer:

K_c\text{ = }\frac{[O_2][H_2]\placeholder{⬚}^2}{[H_2O]\placeholder{⬚}^2}

Explanation:

Here, we want to write the equilibrium constant expression

To write this, we raise the concentrations of the reactants and products to the coefficient before them. These concentrations are represented by square brackets in which the chemical formula of the compound is placed

We place the representation of the products over that of the reactants

We have the expression written as follows:

K_c\text{ =  }\frac{[H_2]\placeholder{⬚}^2[O_2]}{[H_2O]\placeholder{⬚}^2}

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1 year ago
What volume would 3.01•1023 molecules of oxygen gas occupy at STP?
frosja888 [35]
<h3>Answer:</h3>

                  Volume  = 11.2 L

<h3>Explanation:</h3>

Step 1: Calculate Moles:

                      As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

                          Number of Moles  =  Number of Particles ÷ 6.022 × 10²³

Putting values,

                          Number of Moles  =  3.01× 10²³ Particles ÷ 6.022 × 10²³

                          Number of Moles  =  0.50 Moles

Step 2: Calculate Volume:

As we know that one mole of any Ideal gas at standard temperature and pressure occupies exactly 22.4 dm³ volume.

When 1 mole gas occupies 22.4 dm³ at STP then the volume occupied by 0.50 moles of gas is calculated as,

                      = (22.4 dm³ × 0.50 moles) ÷ 1 mole

                      = 11.2 dm³                                       ∴ 1dm³ = 1 L

So,

                                      Volume  = 11.2 L

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Why does a cell make a copy of its DNA before it dvides?
Natasha_Volkova [10]
<span>so there is a set of DNA for the daughter cell after mitosis has occurred</span>
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3 years ago
You are carefully watching the temperature of your melting point apparatus as it is heating up. At 132 C it is still a white sol
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Answer:

See the answer below

Explanation:

<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>

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2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.

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100 POINTS PLEASE HELP!! Honors Stoichiometry Activity Worksheet Instructions: In this laboratory activity, you will taste test
Shtirlitz [24]

Answer:

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=

236.59g/mol

946.36g

=4mol

Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=

225g/mol

196.86g

=0.8749mol

Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=

257.83g/mol

193.37g

=0.7499mol

Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=

719.42g/mol

2050.25g

=2.8498mol

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

\frac{2}{1}\times 0.7499 mol = 1.4998 mol

1

2

×0.7499mol=1.4998mol of water

Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

\frac{1}{1}\times 0.7499 mol = 0.7499 mol

1

1

×0.7499mol=0.7499mol of water

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

\frac{4}{1}\times 0.7499 mol=2.996 mol

1

4

×0.7499mol=2.996mol of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

theoretical yield

Experimental yield

×100

\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%

2157.9722g

2050.25g

×100=95.00%

6 0
3 years ago
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