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Likurg_2 [28]
2 years ago
14

Group VIII corresponds to the...

Chemistry
1 answer:
goldfiish [28.3K]2 years ago
6 0

Answer:

Noble gases farmily

Explanation:

8A = noble gases

This will be on the far right of the periodic table.

- Hope that helped! Please let me know if you need further explanation.

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Calculate the specific heat of a substance when 63j of energy are transferred as heat to an 8.0 g sample to raise it temperature
Flura [38]

The formula for energy or enthalpy is:

E = m Cp (T2 – T1)

where E is energy = 63 J, m is mass = 8 g, Cp is the specific heat, T is temperature

 

63 J = 8 g * Cp * (340 K – 314 K)

<span>Cp = 0.3 J / g K</span>

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What does the 32.06 represent?
lana [24]

The 32.06 represent The atomic mass, the average number of protons & neutrons

<h3>Further explanation </h3>

The Atomic Number (Z) indicates the number of protons in an atom of an element. If the atom is neutral then the number of protons will be equal to the number of electrons. So the atomic number can also indicate the number of electrons.

So atomic number = number of protons = number of electrons

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So that the relationship between atomic numbers and mass numbers can be formulated as follows:

Atomic Number (Z) = Mass Number (A) - Number of Neutrons

In the following element notation,

\large {{{Z} \atop {A}} \right X}

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A = mass number

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Firdavs [7]
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A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu
bogdanovich [222]

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

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The temperature of an object can be written using different temperature scales.

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- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

T(K)=T(^{\circ}C)+273.15

Therefore  in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

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