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Mariulka [41]
3 years ago
5

11. What is the maximum number of electrons that an atomic orbital can contain?

Chemistry
1 answer:
Arada [10]3 years ago
6 0

Answer:

2

Explanation:

Each orbital can hold two electrons. One spin-up and one spin-down.

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Which of these is caused by a chemical change?
IRISSAK [1]
I think that is A is the answer.
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3 years ago
What are some examples of how matter moves through a cycle on earth ​
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Ocean, lakes and rivers. Are all liquids.

Explanation:

Ocean, lakes and rivers. Are all liquids. Snow starts off as a liquid, evaporates into a gas and camoes back as snow.

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3 years ago
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



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4 years ago
Water’s freezing point is 32 degrees F = _____ Celcius = _____ Kelvin
lutik1710 [3]

Waters freezing point in Celsius is 0 and 273.15 Kelvin (or 273 rounded to a whole number.

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