When exothermic reactions occur, _____________

Nylon is an important component in many products. One step in the production of nylon is the reaction of cyclohexane with oxygen

UkoKoshka [18]

Answer:

Follows are the answer to this question:

Explanation:

pleas find the attachment file.

Formula for ΔE:

$\Delta E = \sum \text{Bond energy of recatants}- \sum \text{Bond energy of Products}$

$= 6 ( B.E_{C-C}) +12 (B.E_{C-H}) +(B.E_{o \ = \ o}) - 6(B.E_{C-C})-10(B.E_{C-H}) -( B.E_{C=0}) -2(B.E_{0-H})\\\\ = \bold{2 (B.E_{C-H}) +(B.E_{o \ = \ o}) -( B.E_{C=0}) -2(B.E_{0-H})}\\\\$

to calculate the value for ΔE put the value in above formula:

4 0

3 years ago

SO2+02➡SO3 <br>balance chemical equation

Slav-nsk [51]

Answer:

2S02 + O2 ---> 2SO3

<u><em>Its</em><em> </em><em>balanced</em><em> </em><em>now</em><em> </em><em>as</em><em> </em><em>Reactants</em><em> </em><em>=</em><em> </em><em>Products</em><em>.</em><em> </em></u>

8 0

2 years ago

What passes through the Haversian canal

Tamiku [17]

A Haversian canal is referred to as any minute tube, which forms a network in a bone and also contains blood vessels. Blood vessels, nerve fibers and lymphatics may pass through the Haversian canal. Haversian canals are found surrounding blood vessels and nerve cells all over the bones and interacts with bone cells from end to end networks known as canaliculi.

4 0

3 years ago

Molybdenum can form a wide series of halide compounds, including four different fluoride compounds. The percent by mass of molyb

azamat

The formula and names of the compounds are:

1. Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride

4. Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride

1. Determination of the name and formula of the molybdenum fluoride having 63.0% of molybdenum.

Molybdenum (Mo) = 63.0%

Fluorine (F) = 100 – 63 = 37%

<h3>Formula =? </h3>

Mo = 63.0%

F = 37%

Divide by their molar mass

Mo = 63.0 / 96 = 0.656

F = 37 / 19 = 1.947

Divide by the smallest

Mo = 0.656 / 0.656 = 1

F = 1.947 / 0.656 = 3

Therefore,

Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Determination of the name and formula of the molybdenum fluoride having 56.0% of molybdenum.

Molybdenum (Mo) = 56.0%,

Fluorine (F) = 100 – 56 = 44%

<h3>Formula =? </h3>

Mo = 56%

F = 44%

Divide by their molar mass

Mo = 56 / 96 = 0.583

F = 44 / 19 = 2.316

Divide by the smallest

Mo = 0.583 / 0.583 = 1

F = 2.316 / 0.583 = 4

Therefore,

Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Determination of the name and formula of the molybdenum fluoride having 50.0% of molybdenum.

Molybdenum (Mo) = 50.0%,

Fluorine (F) = 100 – 50 = 50%

<h3>Formula =? </h3>

Mo = 50%

F = 50%

Divide by their molar mass

Mo = 50 / 96 = 0.520

F = 50 / 19 = 2.632

Divide by the smallest

Mo = 0.520 / 0.520 = 1

F = 2.632 / 0.520 = 5

Therefore,

Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride

4. Determination of the name and formula of the molybdenum fluoride having 46.0% of molybdenum.

Molybdenum (Mo) = 46.0%,

Fluorine (F) = 100 – 46 = 54%

<h3>Formula =? </h3>

Mo = 46%

F = 54%

Divide by their molar mass

Mo = 46 / 96 = 0.479

F = 54 / 19 = 2.842

Divide by the smallest

Mo = 0.479 / 0.479 = 1

F = 2.842 / 0.479 = 6

Therefore,

Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride

Learn more: brainly.com/question/11185156

7 0

2 years ago

How much heat in j is giving out when 85.0g of lead cools from 200.0 c to 10.0c

photoshop1234 [79]

Answer:

q = - 2067.2 J of Heat is giving out when 85.0g of lead cools from 200.0 c to 10.0 c.

Explanation:

The Specific Heat capacity of Lead is 0.128 $\frac{J}{g\ ^{0}C}$

This means, increase in temperature of 1 gm of lead by $1 ^{0}\ C$ will require 0.128 J of heat.

Formula Used :

$q = m.c.\Delta T$

q = amount of heat added / removed

m = mass of substance in grams = 85.0 g

c = specific heat of the substance = 0.128

$q = m.c.\Delta T$ = Change in temperature

= final temperature - Initial temperature

= 10 - 200

= - $190 ^{0}\ C$

put value in formula

q = - $85\times 0.128\times 190$

On calculation,

q = - 2067.2 J

- sign indicates that the heat is released in the process

5 0

3 years ago

When exothermic reactions occur, ______________.