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ahrayia [7]
3 years ago
12

In a synthesis reaction, one reactant contains 346 J of chemical energy, and one reactant contains 153 J of chemical energy. The

product contains 435 J of chemical energy. Because energy is conserved, what energy change occurs during the reaction?
Chemistry
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

64J of energy must have been released.

Explanation:

Step 1: Data given

One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.

The product contains 435 J of chemical energy.

Step 2:

Since the energy is conserved

Sum of energy of Reactants = Energy of Products

Sum of energy of Reactants = 346 J + 153 J = 499 J

The energy of the product = 435 J

435 < 499

This means energy must have been lost as heat.

Step 3: Calculate heat released

499 J - 435 J = 64 J

64J of energy must have been released.

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Bonds between carbon and oxygen are more polar than bonds between sulfur and oxygen. nevertheless, sulfur dioxide (SO₂) exhibits a dipole moment while carbon dioxide (CO₂) does not because of the difference in their shape, CO₂ is having linear geometry thus exhibit zero dipole moment while SO₂ is having bent shape thus exhibit dipole moment. So, despite the fact that bonds between carbon and oxygen are more polar than bonds between sulfur and oxygen. nevertheless, sulfur dioxide (SO₂) exhibits a dipole moment while carbon dioxide (CO₂) does not.

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3 years ago
The first-order rearrangement of CH3NC is measured to have a rate constant of 3.61 x 10-15 s-1 at 298 K and a rate constant of 8
laiz [17]

Answer:

The correct answer is 160.37 KJ/mol.

Explanation:

To find the activation energy in the given case, there is a need to use the Arrhenius equation, which is,  

k = Ae^-Ea/RT

k1 = Ae^-Ea/RT1 and k2 = Ae^-Ea/RT2

k2/k1 = e^-Ea/R (1/T2-1/T1)

ln(k2/k1) = Ea/R (1/T1-1/T2)

The values of rate constant k1 and k2 are 3.61 * 10^-15 s^-1 and 8.66 * 10^-7 s^-1.  

The temperatures T1 and T2 are 298 K and 425 K respectively.  

Now by filling the values we get:  

ln (8.66*10^-7/3.61*10^-15) = Ea/R (1/298-1/425)

19.29 = Ea/R * 0.001

Ea = 160.37 KJ/mol

3 0
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A nucleus with four protons has total positive charge
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It would have a charge of 4+
7 0
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A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h
statuscvo [17]

Explanation:

First, calculate the moles of CO_{2} using ideal gas equation as follows.

                PV = nRT

or,          n = \frac{PV}{RT}

                = \frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = \frac{mass}{volume}

Hence, mass of water will be as follows.

                Density = \frac{mass}{volume}

             0.998 g/ml = \frac{mass}{3.26 ml}    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                              =  \frac{3.25 g}{18.02 g/mol}            

                              = 0.180 mol

Moles of hydrogen = 0.180 \times 2 = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

          0.183 mol =  \frac{mass}{12 g/mol}            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                               =  \frac{0.95 g}{16 g/mol}            

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is C_{3}H_{6}O.

Thus, we can conclude that empirical formula of the given compound is C_{3}H_{6}O.            

6 0
3 years ago
student in chemistry 150-02 weighed out 55.5g of octane c8h18 and allowed it to react with oxygen, o2 the product formed were ca
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Answer:

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Explanation:

unbalanced reaction:

C8H8 + O2 → CO2 + H2O

balanced for semireactions:

(1) 16H2O + C8H8 → 8CO2 + 40H+

(2) 10(4H+ + O2 → 2H2O)

⇒ 40H+ + 10O2 → 20H2O

(1) + (2):

balanced reaction:

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                     20 - O2 - 20

                      8 - H - 8

3 0
4 years ago
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