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oksian1 [2.3K]
3 years ago
10

3. If the dartboard below is used to model an atom, which subatomic particles would be located at Z?

Chemistry
1 answer:
yarga [219]3 years ago
3 0

Answer : The protons and neutrons subatomic particles will be located at Z.

Explanation :

In the model of an atom, there are three subatomic particles. Protons, neutrons and electrons are the subatomic particles.

The protons and the neutrons subatomic particles are located inside the nucleus and the electrons subatomic particle are located around or outside the nucleus.

The protons are positively charged, electrons are negatively charged and neutrons are neutral that means it has no charge.

In the given dartboard, Z is the nucleus in which the protons and neutrons subatomic particles are present and x, w & y are the electrons because they are located around the nucleus.

Hence, the protons and neutrons subatomic particles will be located at Z.

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The density of lead is
Mademuasel [1]

Answer : The volume occupied by lead is, 2.20 mL

Explanation :

Density : It is defined as the mass contained per unit volume.

Formula used for density :

Density=\frac{Mass}{Volume}

Given :

Mass of an lead = 25.0 grams

Density of lead = 11.34 g/mL

Now put all the given values in the above formula, we get the volume of lead.

11.34g/mL=\frac{25.0g}{Volume}

Volume=2.20mL

Therefore, the volume occupied by lead is, 2.20 mL

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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo
riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)

The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

\Delta H = heat released by the reaction = ?

m = mass of benzene = 125 g

c_{p,g} = specific heat of gaseous benzene = 1.06J/g^oC

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

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

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

4 0
3 years ago
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