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Mila [183]
3 years ago
11

The circles in the model represent atmospheric layers. Make each description with the correct name of each layer.

Chemistry
1 answer:
Masja [62]3 years ago
6 0

Answer:

The bottom is troposhere the next is sratoshere,mesosphere,thermosphere then exoshere

Explanation:

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Which statement is not true regarding nuclear reactions? Alpha, beta, and gamma are types of radiation. Nuclear reactions involv
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Alpha I believe not sure
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Which planet is terrestrial?<br> Jupiter<br> Mars<br> O Saturn<br> Uranus
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Explanation: The planet of Mars is the only planet of all that is terrestrial. Jupiter, Saturn and Neptune are all Jovian Planets. The answer to the question is Mars. Hope this answer helps you!

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Gold (au) goes from an oxidation number of +3 to 0 in the reaction below. 2Au3+ + 6I- ---&gt; 2Au + 3I2. what describes the Au3+
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In a reduction-oxidation or better known as REDOX reaction, the substance that reduces the oxidation state is known as the substance that is REDUCED. It serves as the oxidizing agent. Thus, Au3+  in this number is considered as the oxidizing agent. 
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The attractions that allow molecules of krypton to exist in the solid phase are due to:
OLga [1]

Answer:

4) Van der waals forces

Explanation:

Krypton (Kr) belongs to the noble gas group and has fully filled valence orbitals. In the solid phase, Kr exists as a white solid with a face centered cubic structure.

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Ionic > hydrogen bonding > dipole-dipole > london dispersion

Kr is monoatomic and non-polar. When fully filled (stable) valence orbitals of 2 Kr atoms approach each other in close proximity they experience a repulsive force which prevents the formation of strong bonds. Thus, the only force of attraction in Kr is the long range weak Van Der Waals force also known as the london dispersion force.

7 0
3 years ago
Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
3 years ago
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