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BaLLatris [955]
3 years ago
13

At atom whose valence electron shell is nearly full is ________ chemically reactive. a. not b. mildly c. highly d. the name of t

he element is needed to figure this out.
Physics
2 answers:
svet-max [94.6K]3 years ago
5 0
I believe the correct response would be B. Mildly reactive. As the atom simply needs only a few more electrons in its valence shell to achieve stability and have an octet of 8 valence electrons in its valence shell.
Vaselesa [24]3 years ago
4 0
I bet is a because it will not be chemically
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You push a shopping cart for fun with a force of 60N. If the shopping cart has a mass of 12kg, what is the
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Answer:

\huge\boxed{\sf a = 5 \ ms^{-2}}

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<u>Solution:</u>

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a = F / m

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\rule[225]{225}{2}

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Which of these is an example of physical weathering? Rock breaking up due to carbonic acid Rock breaking up due to mechanical fo
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Biological weathering

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3 years ago
While John is traveling along an interstate highway, he notices a 150 mi marker as he passes through town. Later John passes a 1
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Part a)

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Part b)

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7 0
4 years ago
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A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
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Answer:

A. 1.64 J

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First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

4 0
3 years ago
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