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hichkok12 [17]
4 years ago
7

Why are the atoms of the group 8 elements unreactive??​

Chemistry
1 answer:
anygoal [31]4 years ago
3 0

Answer:

because that is the last group the ones before it are more reactive im sorry if this not what you were looking for.

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Which of the following best describes the uses of private land?
melamori03 [73]

Answer : (C) It can be used for variety of activities and 4es

Explanation: A private land is generally used for any purpose like home, business, office, factories or any other purpose. The land is privately owned by its owner which can use the private land for variety of activities the owner wants. A private land can be used as a farm land, or any other non agriculture purpose like building a factory, converting into home or any other purpose.

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3 years ago
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3. The drop factor (gtts/cc) is the number of drops it takes to equal 1 cc for specific tubing in intravenous fluids. All
viva [34]

Answer:

30gtts/min

Explanation:

To solve this problem we need to divide the volume(mL) and time(min) then multiply by the drop factor. Note that 1hr = 60min.

120/60 * 15 = 30gtts/min

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7 0
3 years ago
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mestny [16]

Answer:

A

Explanation:

it causes the genes of two different individuals to mix

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3 years ago
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Based on your understanding of how bond types influence a material’s properties, identify each of the following compounds as bei
vfiekz [6]

Answer:

Steel:   metallic

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Calcium chloride:   ionic

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Explanation:

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3 years ago
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Sabendo que os calores de combustão do enxofre monoclínico e do enxofre rômbico são, respectivamente, - 297,2 kJ/mol e - 296,8 k
liq [111]

Responda:

+ 0,9kJ / mol

Explicação:

Dados os calores de combustão do enxofre monoclínico e enxofre rômbico como - 297,2 kJ / mol e - 296,8 kJ / mol, respectivamente para a variação na transformação de 1 mol de enxofre rômbico em enxofre monoclínico conforme mostrado pela equação;

S (mon.) + O2 (g) -> SO2 (g)

Uma vez que são todos 1 mol cada, a mudança na entalpia será expressa como ∆H = ∆H2-∆H1

Dado ∆H2 = -296,8kJ / mol

∆H1 = -297,2kJ / mol

∆H = -296,8 - (- 297,2)

∆H = -296,8 + 297,2

∆H = 297,2-296,8

∆H = + 0,9kJ / mol

Portanto, a mudança na entalpia da equação é + 0,9kJ / mol

4 0
3 years ago
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