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ser-zykov [4K]
3 years ago
12

Picking up a hot coal is an example of what type of heat transfer?

Chemistry
2 answers:
Pavlova-9 [17]3 years ago
5 0
Conduction should be the correct answer.
expeople1 [14]3 years ago
4 0

C. Conduction.


try and good luck. :)

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Pls help Would be much appreciated:)
Pie

Answer:

Ok so,  b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...

i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced

ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.

iii. Calculate the standard potential (voltage) of the cell

Look up the reduction potential,

E

⁰

red

, for the reduction half-reaction in a table of reduction potentials

Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,

E

⁰

ox

=

-

E

⁰

red

.

iv. What kind of electrochemical cell is this? Explain your answer.

All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells

1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery

Explanation:

3 0
3 years ago
Nitrogen monoxide, NO, is formed in automobile exhaust by the reaction of nitrogen and oxygen in the air: N2 (g) + O2 (g) ↔ NO (
Solnce55 [7]

Answer:

The equilibrium concentration of NO is 0.001335 M

Explanation:

Step 1: Data given

The equilibrium constant Kc is 0.0025 at 2127 °C

An equilibrium mixture contains 0.023M N2 and 0.031 M O2,

Step 2: The balanced equation

N2(g) + O2(g) ↔ 2NO(g)

Step 3:  Concentration at the equilibrium

[N2] = 0.023 M

[O2] = 0.031 M

Kc = 0.0025 = [NO]² / [N2][O2]

Kc = 0.0025 = [NO]² / (0.023)(0.031)

[NO] = 0.001335 M

The equilibrium concentration of NO is 0.001335 M

6 0
3 years ago
Equator runs east and west all the way around the world,halfway between the south poles called the ______________and north poles
Korvikt [17]

Answer:

hanigin ng mga pule natin

3 0
3 years ago
What characteristics of a metallic bond explains some of the properties of metals
d1i1m1o1n [39]

Answer:

See Explanation

Explanation:

Metallic bonds involve attraction between electrons and positively charged metal ions. The metals are ionized and electrons form a sea of valence electrons. These loosely bound electrons surround the nuclei of the metals.

The presence of this sea of electrons explains the fact that metals conduct electricity and heat due to the free valence electrons.

Due to the nature of the bonding between metal atoms,metals are malleable and ductile.

Due to the strong electrostatic interaction between metal ions and electrons, the metallic bond is very strong and is very difficult to break thereby accounting for the greater strength of metals as the size of the metallic ion decreases.

8 0
3 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
3 years ago
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