A hydrogen fuel cell is an electrochemical reactor in which oxygen and hydrogen gas are reacted to produce energy. Rather than combusting hydrogen in the presence of oxygen, the overall reaction (1/2) O2 H2 → H2O is actually carried out as two `half-reactions’ occurring at separate electrodes. To complete the reaction, this forces electrons to pass through an external circuit from one electrode to the other, and this current can be used to do work; e.g., by driving a motor. Typical operating conditions for a solid-oxide fuel cell are 500-900 C and 5-15 bar. Consider a fuel cell in which a pure oxygen stream and a pure hydrogen stream enter at 11 bar and 700 C with stoichiometric flowrates. Assume that these reactants are consumed completely, and that the water generated by the reaction leaves in a third stream at the same T and P. Assume that the fuel cell operates at steady-state with a uniform temperature of 700 C. a) Write simplified first and

Answer:

trigonal planar

Explanation:

Answer:

1. to latch onto and let go of eggs

2. how quickly electrical energy moves in a circuit

3.  when a current runs through a conductor

4. Electrons are moved through a circuit, creating a current. That current interacts with a magnet, moving the conductor in the orientation of the field.

5.  electromagnetic field

Answer:

what grade is this?

Answer:

7. Option D

Explanation:

Solubility increases with the increase in temperature of the solution. It is so because the increase in temperature leads to increase in kinetic energy which then breaks the solute particles held together by intermolecular attractions.

Hence, option D is correct

Answer:

A via A P E X

The entity that would make ΔG to become negative at a given enthalpy and entropy is Changing the temperature (Option A)

ΔG = ΔH – TΔS

Where

NOTE

ΔG = +ve (non spontaneous)

ΔG = 0 (equilibrium)

ΔG = –ve (spontaneous)

From the Gibbs free energy equation, we can see that the entities that makes us the equation are

Thus, we can conclude that the correct answer to the question is: changing the temperature (Option A) since the equation has nothing to do with neither concentration, pressure nor volume

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Answer:

I honestly don't know but I hope you get more points! ;D

Explanation:

Answer:

Explanation:

c it’s see it

Answer:

See explanation

Explanation:

a) The magnitude of intermolecular forces in compounds affects the boiling points of the compound. Neon has London dispersion forces as the only intermolecular forces operating in the substance while HF has dipole dipole interaction and strong hydrogen bonds operating in the molecule hence HF exhibits a much higher boiling point than Ne though they have similar molecular masses.

b) The boiling points of the halogen halides are much higher than that of the noble gases because the halogen halides have much higher molecular masses and stronger intermolecular forces between molecules compared to the noble gases.

Also, the change in boiling point of the hydrogen halides is much more marked(decreases rapidly)  due to decrease in the magnitude of hydrogen bonding from HF to HI. The boiling point of the noble gases increases rapidly down the group as the molecular mass of the gases increases.

Answer:

The optical properties of a material define how it interacts with light. The optical properties of matter are studied in optical physics, a subfield of optics. Wiki.pedia

Explanation:

Answer:

true

Explanation:

Answer:

To keep from falling apart

Explanation:

Answer:

Reaction that takes place at the Zn electrode is oxidation:

Zn ⇄  Zn²⁺  +  2e⁻

Explanation:

This excersise explains the structure of a typical battery, that is useful to study electrochemistry.

AgNO₃ → Ag⁺ + NO₃

ZnSO₄ → Zn²⁺  +  SO₄⁻²

Generally batteries contains 2 electrodes, where we have 2 conductive metallic. Batteries work to obtain electricity from a chemistry reaction which is always a redox type. One electrode releases electrons while the other catch them. Electrons travel from the anode to cathode, so in the anode, the e⁻ are released.

Reaction that takes place at the Zn electrode is oxidation:

Zn ⇄  Zn²⁺  +  2e⁻

Oxidation state is increased, electrons are free to go to cathode, where the other element decreases the oxidation state:

Ag⁺ + e⁻  ⇄ Ag

In the middle of the battery, the salt bridge (generally KCl) contributes the redox with more charges. Cl⁻ for the anode and K⁺ for the cathode.

Note that the same amount of electrons travel from anode to cathode, because in this example, Ag gained 1 mol and Zn released 2 moles.

The scheme for the battery is:

⁻  Zn(s) / Zn²⁺ (aq) //  Ag⁺ (aq) / Ag (s) ⁺

         1 M                               1 M                      

Where the first place represents the anode, // represents the salt bridge and then, the cathode.  Below, we write the molar concentration of each salt.

the answer is gold cause oil wood and glass are electrical insulators

Shaking causes small bubbles, which helps the soda's carbon dioxide escape faster. The gas will gradually escape from the liquid as bubbles once the can is opened, and the soda will become "flat." The dissolved gas takes a long time to escape if the liquid is treated gently.

Answer:

[tex]1s^2\, 2s^2[/tex].

Explanation:

Electron orbitals in an atom (e.g., [tex]1s[/tex]) are denoted with:

There are two aspects to consider when finding the electron configuration of an atom:

The [tex]s[/tex] orbital in each shell could hold up to [tex]2 \times 1 = 2[/tex] electrons (one [tex]s\![/tex] orbital per shell, with up to two electrons.)

The [tex]p[/tex] orbitals in each shell could hold up to [tex]2 \times 3 = 6[/tex] electrons (three [tex]p\![/tex] orbitals per shell, with up to two electrons in each orbital.)

The [tex]d[/tex] orbitals in each main shell could hold up to [tex]2 \times 5 = 10[/tex] electrons (five [tex]d\![/tex] orbitals per shell, with up to two electrons in each orbital.)

Refer to the order in which the orbitals are filled (Aufbau principle.)

All four electrons of Beryllium are thus assigned to the [tex]1s[/tex] and [tex]2s[/tex] orbitals. In a ground-state Beryllium atom, orbitals [tex]2p[/tex] and beyond would contain no electrons.

Notation:

Write the non-empty orbitals in the order by which they are filled:

[tex]1s^2\, 2s^2[/tex].

Answer:

Answer: pH = 2.72

Explanation:

Calculate the pH of 0.010 M HNO2 solution. The K, for HNO2 is 4.6 x 104 

Answer:

Hello your question lacks some required information attached below is the missing information

answer :

number of moles

Acetic acid = 0.061199 mole

Alcohol = 0.027320 mole

Explanation:

The starting volumes :  2.5mL of either alcohol and 3.5mL of either acid

for the purpose of this solution I will take 2.5 mL of 1-butanol and 3.5 mL of acetic acid

mass of 1-Butanol = 2.5 mL * 0.81 g/mol = 2.025 g

∴ number of moles of 1-butanol = 2.025g / 74.12 g/mol = 0.027320 mole

mass of acetic acid = 3.5 mL * 1.05 g/mol = 3.675 g

∴ number of moles of Acetic acid = 3.675 g/ 60.05 g/mol = 0.061199 mole

Reaction equation

CH3-CH2-CH2-CH2-OH + CH3-COOH + H+ (cat) ↔ CH3-CH2-CH2-CH2-O-CO-CH3 + H2O

Answer:

[tex]0.769\ \text{M}[/tex]

Explanation:

Mass of stock solution = 20 g

Molar mass of NaOH = 40 g/mol

Volume of stock solution = 0.150 mL

[tex]M_2[/tex] = Concentration of NaOH for the final solution

[tex]V_1[/tex] = Amount of stock solution taken = 15 mL

[tex]V_2[/tex] = Total volume of solution = 65 mL

Molarity is given by

[tex]M_1=\dfrac{\text{Mass}}{\text{Molar mass}\times \text{Volume}}\\\Rightarrow M_1=\dfrac{20}{40\times 0.15}\\\Rightarrow M_1=\dfrac{10}{3}[/tex]

We have the relation

[tex]M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{\dfrac{10}{3}\times 15}{65}\\\Rightarrow M_2=0.769\ \text{M}[/tex]

The concentration of NaOH for the final solution is [tex]0.769\ \text{M}[/tex].

Answer: Animals found in the Arctic tundra include herbivorous mammals (lemmings, voles, caribou, arctic hares, and squirrels), carnivorous mammals (arctic foxes, wolves, and polar bears), fish (cod, flatfish, salmon, and trout), insects (mosquitoes, flies, moths, grasshoppers, and blackflies), and birds (ravens, snow buntings

The lives in Artic tundra biome is caribou.

Artic tundra biome is the northernmost biome, which covers the land of the arctics with the ice caps.

Climate of this Artic tundra biome is almost cold means the temperature in winter season of this region is about -34 degree celsius. In this region small animals like Norway lemmings as well as large animals like caribou are present which have high amount of fat to rescue from the cold.

Hence, option (a) is correct.

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Answer: 49.8 g

Explanation: molar mass M(KI) = 39.1+ 126.9 = 166 g/mol

Mass m= n·M= 0.3 mol· 166 g/mol

Answer:

496 g of Fe₂O₃.

Explanation:

The balanced equation for the reaction is given below:

4Fe + 3O₂ —> 2Fe₂O₃

From the balanced equation above,

4 moles of Fe reacted to produce 2 moles of Fe₂O₃.

Therefore, 6.20 moles of Fe will react to produce = (6.20 × 2)/4 = 3.1 moles of Fe₂O₃

Finally, we shall determine the mass of 3.1 moles of Fe₂O₃. This can be obtained as follow:

Mole of Fe₂O₃ = 3.1 moles

Molar mass of Fe₂O₃ = (56 × 2) + (3×16)

= 112 + 48

= 160 g/mol

Mass of Fe₂O₃ =?

Mass = mole × molar mass

Mass of Fe₂O₃ = 3.1 × 160

Mass of Fe₂O₃ = 496 g

Therefore, 496 g of Fe₂O₃ were produced from the reaction.

Answer:

A) atomic size

Explanation:

Answer:

140 mol CO₂.

Explanation:

In order to convert from moles of octane (C₈H₁₈) into moles of CO₂, we can use a conversion factor with the stoichiometric coefficients of the balanced reaction:

140 moles of CO₂ can be produced from 17.5 moles of octane.

Answer:

3 × 10⁴ kJ

Explanation:

Step 1: Write the balanced thermochemical equation

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ

Step 2: Calculate the moles corresponding to 865.9 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

865.9 g × 1 mol/18.02 g = 48.05 mol

Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced

According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.

48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ

This question is incomplete, the complete question is;

g Consider a pure sample of a radioactive isotope with a mass number of (46 + A). If the sample has mass of (25.0 + B) micrograms and the isotope has a half-life of (4.50 + C)x10⁶ years,

determine the decay rate for the sample. Give your answer in decays/second and with 3 significant figures.

A = 1, B = 1, C = 11.

Answer:

the decay rate for the sample is 469,625.898 decay/s    

Explanation:

Given the data in the question;

[tex]T_{\frac{1}{2}[/tex] = ( 4.50 + C ) × 10⁶ years  

C = 11

[tex]T_{\frac{1}{2}[/tex] = ( 4.50 + 11 ) × 10⁶ years

[tex]T_{\frac{1}{2}[/tex] = 15.50 × 10⁶ years

[tex]T_{\frac{1}{2}[/tex] = 15.50 × 10⁶ × 365 × 24 × 60 × 60  

λ = 0.693 / [tex]T_{\frac{1}{2}[/tex]

so

λ = 0.693 / ( 15.50 × 10⁶ × 365 × 24 × 60 × 60  )

also

N = ( 25.0 + B )

B = 1

N = ( 25.0 + 1 )

N = 26 × 10⁻⁶

isotope with a mass number of ( 46.0 + A ) , A=1

= ( 46.0 + 1 ) = 47

N = 26 × 10⁻⁶( 1/47 )( 1 / 1.67 × 10⁻²⁷ ) =  3.31252 × 10²⁰  

R = λN

so

R = (0.693 / ( 15.50 × 10⁶ × 365 × 24 × 60 × 60  )) × (3.31252 × 10²⁰)

R = 1.41773 × 10⁻¹⁵ × (3.31252 × 10²⁰)

R = 469,625.898 decay/s    

Therefore, the decay rate for the sample is 469,625.898 decay/s