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3 years ago

Match the atoms to their correct electron configuration. (Answer choices in photo)

Chemistry

1 answer:

zysi [14]3 years ago

6 0

Answer:

The answer to your questions are given below

Explanation:

The answer to the questions given above can simply be obtained by writing the electronic configuration of each atom.

The electronic configuration of each atoms can be written as follow:

Fluorine, F (9) => 1s² 2s²2p⁵

Sodium, Na (11) => [Ne] 3s¹

Helium, He (2) => 1s²

Calcium, Ca (20) => [Ar] s²

Nitrogen, N (7) => 1s² 2s²2p³

Sulphur, S (16) => [Ne] 3s²3p⁴

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What is the answer to-6 + 1 = -5/

Jlenok [28]

-6+1= -5
-5 is the true answer

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3 years ago

A 8 kg cat is running 4 m/s. How much kinetic energy does it have

Debora [2.8K]

<span>KE = 1/2mv^2
KE = 1/2(8)4 m/s^2
KE = 4*4
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3 years ago

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

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

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

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3 years ago

How many joules of energy do you have if you have 0.45 L of water in a can and you heat the 23C water to 55C site:socratic.org

Bess [88]

Answer:

Q = 60192 j

Explanation:

Given data:

Volume of water = 0.45 L

Initial temperature = 23°C

Final temperature = 55°C

Amount of heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 55°C - 23°C

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one L = 1000 g

0.45 × 1000 = 450 g

Specific heat capacity of water is 4.18 j/g°C

Q = m.c. ΔT

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Q = 60192 j

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3 years ago

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Andrews [41]

Answer: I agree with the student because in the question prior to this One question stated thatExtensive properties very with the amount of matter ina sample, so yes i agree.

Explanation:

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3 years ago