C4H9OH(g)+O2(g)→CO2(g)+?H2O(l) what coefficient should be placed in front of H2O to balance the hydrogen atoms?

Chemistry

1 answer:

slava [35]3 years ago

3 0

Answer:

Explanation:

To balance the hydrogen atoms, we check the number of hydrogen on the left side, this is equal to the 10 hydrogen atoms we have in the alkanol.

Now, on the right hand side, we can see we only have two hydrogen atoms in the water molecule. Now, to make equal the number of hydrogen atoms on both sides, we simply multiply the number of hydrogen there by 5 to make it 10 too

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An atom in an excited state contains more of what type of energy than the same atom in the ground state? An atom in an excited s

kupik [55]

Answer: An atom in an excited state contains more of kinetic energy than the same atom in the ground state.

Explanation:

Kinetic energy is the energy acquired by an object due to its motion. And, thermal energy is the internal energy of an object arisen because of the kinetic energy present within the molecules of the object.

Potential energy is the energy acquired by an object due to its position.

The total energy present at the center of mass of an object is known as mass-energy.

So, when an atom gets excited then it means it is gaining kinetic energy due to which it moves from its initial position after getting excited.

Thus, we can conclude that an atom in an excited state contains more of kinetic energy than the same atom in the ground state.

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3 years ago

What is a way of transforming energy by exerting a force over distance?

xz_007 [3.2K]

Answer:

a. Work

Explanation:

If you apply a force over a given distance - you have done work. Work = Change in Energy. If an object's kinetic energy or gravitational potential energy changes, then work is done. The force can act in the same direction of motion.

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3 years ago

Read 2 more answers

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$