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3 years ago

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A galvanic cell consists of a Al3 /Al half-cell and a standard hydrogen electrode. If the Al3 /Al half-cell standard cell functi

ons as the anode, and the standard cell potential is 1.66 V, what is the standard reduction potential for the Al3 /Al half-cell

1 answer:

11111nata11111 [884]3 years ago

3 0

Answer:

Explanation:

here is your answer:

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Which element has a gas notation of [Kr]5s2 4d7

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I think it would be Kriptonite

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Why do atoms share electrons in covalent bonds??

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Atoms in covalent bonds do combine so as to be stable. As covalent bond consist non metals e.g O2 in this example each atom has vacance of 2 orbitals/ electrons so shairing electrons result their stability

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A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

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3 years ago

Ice must absorb heat to melt into water. What must water do to evaporate?

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Increase in heat, molecules start to escape and it turns to vapor

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When a mercury-202 nucleus is bombarded with a neutron, a proton is ejected. What element is formed?

Galina-37 [17]

That will make a gold-202 nucleus.

<h3>Explanation</h3>

Refer to a periodic table. The atomic number of mercury Hg is 80.

Step One: Bombard the $\displaystyle ^{202}_{\phantom{2}80}\text{Hg}$ with a neutron $^{1}_{0}n$ . The neutron will add 1 to the mass number 202 of $^{202}_{\phantom{2}80}\text{Hg}$ . However, the atomic number will stay the same.

New mass number: 202 + 1 = 203.
Atomic number is still 80.

$^{202}_{\phantom{2}80}\text{Hg} + ^{1}_{0}n \to ^{203}_{\phantom{2}80}\text{Hg}$ .

Double check the equation:

Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.

Step Two: The $^{203}_{\phantom{2}80}\text{Hg}$ nucleus loses a proton $^{1}_{1}p$ . Both the mass number 203 and the atomic number will decrease by 1.

New mass number: 203 - 1 = 202.
New atomic number: 80 - 1 = 79.

Refer to a periodic table. What's the element with atomic number 79? Gold Au.

$^{203}_{\phantom{2}80}\text{Hg} \to ^{202}_{\phantom{2}79}\text{Au} + ^{1}_{1}p$ .

Double check the equation:

Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.

A gold-202 nucleus is formed.

6 0

3 years ago