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KiRa [710]
3 years ago
10

BRAINLIESTTT ASAP! PLEASE ANSWER IN YOUR OWN WORDS. :)

Chemistry
1 answer:
const2013 [10]3 years ago
8 0

Answer:

The atomic structure of an atom involves 3 subatomic particles: the proton, neutron, and electron. The proton has a positive charge and is found in the core of the atom, with the neutral neutrons that also have a mass of 1 amu (atomic mass unit) just like the proton. The nucleus is the core of the atom and contains protons and neutrons and is practically the only area with mass. The electron cloud is basically an area surrounding the nucleus and it contains negative charged electrons. Electrons have no mass but are charged with a negative charge that keeps them. I really hope this helps :)

Explanation:

There is a helpful video that actually explains the structure of an atom in a rather fun way in just 2 minutes. It really does help big time and it's kinda funny if you look it up on YT and watch:

WKRP: Venus Explains the Atom

Have a wonderful great day :)

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3 years ago
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The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat pr
jenyasd209 [6]

Answer : The final temperature of the water is, 22.5166^oC

Explanation : Given,

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First we have to calculate the moles of benzene.

\text{Moles of benzene}=\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{8.600g}{78g/mol}=0.1103mole

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The given balanced chemical reaction is:

2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ

According to reaction,

As, 2 moles of benzene gives 6542 kJ of energy on combustion.

So, 0.1103 mole of benzene gives \frac{6542 kJ}{2}\times 0.1103=360.7913kJ of energy on combustion.

Now we have to calculate the final temperature of the water.

Formula used : q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})

where,

q_w = heat released = 360.7913 kJ = 36079.13 J

m_w = mass of water = 5691 g

c_w = specific heat of water= 4.18J/g^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 21^oC

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

36079.13J=5691g\times 4.18J/g^oC\times (T_{final}-21^oC)

T_{final}=22.5166^oC

Therefore, the final temperature of the water is, 22.5166^oC

8 0
4 years ago
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500×10-3what about 2?

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