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4 years ago

In an exothermic reaction, energy may be released to the surroundings in the form of?

Chemistry

1 answer:

djverab [1.8K]4 years ago

7 0

Answer:

All of the above

Explanation:

Exothermic reactions release energy usually in the form of heat, but also in a form of light (e.g. a spark, flame, or flash), electricity (e.g. a battery), or sound (e.g. explosion heard when burning hydrogen)

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A gas at 20.0°C and 52.1 mL is heated to 93.5°C. What is the new volume?

lawyer [7]

$\frac{ V_{1} }{ T_{1} }$ = $\frac{ V_{2} }{ T_{2} }$
$V_{1}$ = 0.0521 L
$T_{1}$ = 293.15 K
$T_{2}$ = 366.65 K

Solve for $V_{2}$

$\frac{0.0521 L}{293.15 K}$ = $\frac { V_{2} }{366.65 K}$

$V_{2}$ = 0.06516 L or 65.2 mL

7 0

3 years ago

Which device uses on a non-renewable energy saurce?

aivan3 [116]

Nonrenewable energy resources include coal, natural gas, oil, and nuclear energy. Once these resources are used up, they cannot be replaced, which is a major problem for humanity as we are currently dependent on them to supply most of our energy needs

5 0

3 years ago

C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w

Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃ ΔH = –183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂ ΔH = 183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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#SPJ1

7 0

2 years ago

. Un cuerpo irregular se introduce dentro de una probeta que inicialmente tiene un volumen de 20 ml como muestra la figura Despu

MrMuchimi

Answer:

El volumen del cuerpo es el mismo al comienzo de la experiencia.

Explicación:

El volumen del cuerpo es el mismo al principio porque el volumen no cambia si la temperatura permanece igual. Si cambiamos la temperatura i. mi. Al aumentar la temperatura, las moléculas comienzan a expandirse y se produce un aumento de volumen mientras que cuando disminuimos la temperatura, las moléculas de esa sustancia comienzan a contraerse y el volumen de esa sustancia disminuye. Entonces concluimos que el volumen depende de la temperatura.

6 0

3 years ago

A flask contains 2.0 mol of He gas at 25°C and 1.00 atm. How much He gas, in grams, must be added to increase the pressure to 2.

katrin [286]

Answer:

Mass of He required = 8.0 g

Explanation:

Given,

Initial moles of He = 2.0 mol

Initial pressure = 1.00 atm

final pressure = 2.00 atm

Ideal gas equation,

PV = nRT

As V, R and T are constant

So, $\frac{P_1}{P_2} =\frac{n_1}{n_2}$

$\frac{1.00 atm}{2.00 atm} =\frac{2.0}{n_2}\\n_2=\frac{2.0\times 2.00}{1.00} \\n_2=4.0 mol$

Molar mass of He = 4.00 g/mol

No. of moles of He needs to be added = 4.0 - 2.0 = 2.0 mol

Mass = No. of mole × Molar mass

= 2.0 × 4.0

= 8.0 g

3 0

4 years ago