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3 years ago

An atom has the following electron configuration 1s2 2s2 2p6 3s2 3p4 . How many valence electrons does this Atom have

Chemistry

1 answer:

yawa3891 [41]3 years ago

6 0

Answer:

Explanation:

This atom is sulfur (if the electrons are equal to the protons/not an ion). You can tell the number of valence electrons by looking at the individual shell. The first shell (1s) can only hold 2 electrons. The second shell (2s and 2p) can hold 8 electrons. The third shell (3s and 3p), which is the valence shell, only has 6 out of its possible 8 electrons, so this atom has 6 valence electrons.

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Is a girl doing any work while she stands still and holds a box? Why or why not?

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Yes because she is holding the weight of the box.

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3 years ago

you have been observing an insect that defends itself from enemies by secreting a caustic liquid. analysis of the liquid shows i

Alexus [3.1K]

pH of the buffer solution is 1.76.

Chemical dissociation of formic acid in the water:

HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)

The solution of formic acid and formate ions is a buffer.

[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions

[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate

[HCOOH] = 1.45 M - 0.015 M

[HCOOH] = 1.435 M; equilibrium concentration of formic acid

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pKa = -log 1.8×10⁻⁴ M

pKa = 3.74

Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)

pH = 3.74 + log (0.015 M/1.435 M)

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pH = 1.76

More about buffer: brainly.com/question/4177791

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1 year ago

Select all that apply.

Ilya [14]

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3 years ago

Read 2 more answers

Why was there so much disagreements among the ancient greeks about the nature of the atoms?

olga2289 [7]

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4 0

3 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

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3 years ago