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34kurt
3 years ago
5

Use the Earth's Energy Budget diagram to answer the question. Based on the diagram, less solar energy that reaches Earth is ____

_ than is _____.
A.


reflected, radiated


B.


absorbed, reflected


C.


absorbed, radiated


D.


radiated, reflected
Chemistry
1 answer:
Ivan3 years ago
5 0
<span>absorbed, radiated
Hope this helps. </span>
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How many grams of aluminum will be deposited by 0.1F? (Al=27) a.0.3g b. 0.9. c. 9.0g. d. 2.7g<br>​
andrezito [222]

Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.

<h3>What is electrolysis?</h3>

Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.

The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.

Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium

0.1 F will discharge = 0.1/3 × 27 g of aluminium

mass of aluminium deposited = 0.9 g of aluminium.

Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.

Learn more about electrolysis at: brainly.com/question/26050361

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2 years ago
Choose the nonmetallic elements from the list. Check all that apply yttrium: oxygen: boron: polonium: argon: gallium: carbon:
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Answer:

B, E, & G

Oxygen, Argon, & Carbon

7 0
3 years ago
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Anna [14]
The best answer choice here would be 'Combination'
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4 years ago
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The table shows the number of charged subatomic particles in an ion.
LenKa [72]

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The ion will repel the substance because it has more protons than electrons.

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A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C
Mamont248 [21]

Answer:

T_{eq}=23.85^oC

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

\Delta H_{Cu}=-\Delta H_{H_2O}

Therefore the equilibrium temperature shows up as:

m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC

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