What happens in the following cases. [ ( write two eg for each also write is there any ex

A voltaic cell has a zinc anode and a copper cathode. They are connected by a wire but no salt bridge. What can you predict will

Allisa [31]

The prediction is that B. The electrons will flow to the zinc anode where a negative charge will build up and eventually halt the reaction.

This is known as a chemical element, of the periodic table, that is essential to life and is one of the most widely used metals. Zinc is of considerable commercial importance.

Without the salt bridge, positive and negative charges will build up around the electrodes causing the reaction to stop.

Hence, we know that the purpose of the salt bridge is to keep the solutions electrically neutral and allow the free flow of ions from one cell to another.

Read more about<em> zinc</em> here:

brainly.com/question/25764334

#SPJ1

4 0

2 years ago

The element silver exists in nature as two isotopes: 107Ag has a mass of 106.9051 u, and 109Ag has a mass of 108.9048 u. The ave

Lilit [14]

Answer:

107Ag has abundance of 51.7%

109Ag has abundance of 48.3%

Explanation: Please see attachment for explanation

3 0

3 years ago

What is the atomic mass number of an isotope that has an atomic number of 5 and includes 6 neutrons?

irinina [24]

The answer should be; 11

The atomic mass number is found by combining the number of protons and neutrons

Hope this helps :)

8 0

2 years ago

Read 2 more answers

In the Bohr model of the hydrogen atom, the energy required to excite an electron from n = 2 to n = 3 is _______________ the ene

yawa3891 [41]

Answer:

greater than

Explanation:

4 0

2 years ago

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

4 0

2 years ago