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3 years ago

How many inner, outer, and valence electrons are present in an atom of each of the following elements?

Chemistry

1 answer:

telo118 [61]3 years ago

5 0

Answer:

(a) O = Valance Electrons (6), Inner electrons (8)

(b) Sn = Valance Electrons (2), Inner electrons (36)

(c) Ca = Valance Electrons (2), Inner electrons (20)

(d) Fe = Valance Electrons (2), Inner electrons (26)

(e) Se = Valance Electrons (6), Inner electrons (34)

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Use the picture to answer the question.

natima [27]

Answer:

3 holes

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hard shiny surface

Explanation:

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3 years ago

How do transitional fossils fit with the theory of evolution

iren2701 [21]

These transitional fossils serve as evidence to Darwin's Theory of Natural Selection as conserved traits are proven to survive the test of time. This also supports evolution in a way that is shows gradual change in traits of species over generations due to change in living conditions.

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3 years ago

explain the following observation. When 3-methylbutan-2-ol is treated with HBr, a single alkyl bromide is isolated, resulting fr

babunello [35]

Answer:

1. For the first reaction, there is Carbonation and rearrangement

2. For the second reaction, No Carbonation, No rearrangement.

Explanation:

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3 years ago

A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given

m_a_m_a [10]

Answer:

$A=69.8\frac{bar}{m^6}\\\\ B=0.00302bar*m^6$

2. $W=-8.2kPa$

Explanation:

Hello,

1. In this case, for the given p-V equation, one could use the two states to form a 2x2 linear system of equations in terms of A and B:

$\left \{ {{0.1^2A+0.1^{-2}B=1} \atop {0.04^2A+0.04^{-2}B=2}} \right.$

$\left \{ {{0.01A+100B=1} \atop {0.0016A+625B=2}} \right$

Whose solution by any method for solving 2x2 linear system of equations (elimination, reduction or substitution) is:

$A=69.8\frac{bar}{m^6}\\\\ B=0.00302bar*m^6$

2. Now, for us to compute the work, we must first compute n, as the power relating the pressure and volume for this process:

$P_1V_1^n=P_2V_2^n\\\\\frac{P_1}{P_2}=(\frac{V_2}{V_1} )^n\\\\\frac{1bar}{2bar}= (\frac{0.04m^3}{0.1m^3} )^n\\\\0.5=0.4^n\\\\n=\frac{ln(0.5)}{ln(0.4)} =0.7565$

Now, we compute the work:

$W=\frac{P_2V_2-P_1V_1}{1-n} =\frac{2bar*0.04m^3-1bar*0.1m^3}{1-0.7565} \\\\W=-0.082bar*m^3*\frac{1x10^2kPa}{1bar}\\ \\W=-8.2kPa$

Regards.

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4 years ago

1. A student is examining a cell through a microscope. She notes the

Hoochie [10]

C

Animal cells do not have cell walls or chloroplasts!

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3 years ago