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AnnZ [28]
2 years ago
14

How many electrons will be found in the “p” orbitals in the ground state of chlorine atom?

Chemistry
1 answer:
emmasim [6.3K]2 years ago
7 0

Chlorine, Cl , is located in period 3, group 17 of the periodic table, and has an atomic number equal to 17 . This tells you that a neutral chlorine atom will have a total of 17 electrons surrounding its nucleus. Now, notice that the first energy level doesn't not contain a p-subshell,

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Solutions are said to be what kind of mixture? A) Pure substance B) Heterogeneous C) Homogenous D) Element
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Solutions are said to be C. homogeneous mixtures, composed of two or more substances. It is usually liquid, however it may be solid or gas.
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3 years ago
If temperature is held constant for an ideal gas, as P and V change for a given gas sample, the PV of products will_____. increa
vodka [1.7K]

Answer:

PV=nRT

Explanation:

V=<u>R</u><u>T</u><u>n</u>

P

rearrangement gives

  • PV=nRT
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nT

where P=pressure

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3 0
3 years ago
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(20 points) i) An absorption intensity of 1.00 (in arbitrary units) is observed for the maximum peak of the 1:2:1 triplet of the
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Answer:

Concentration of ethanol required =  48.476 M

Explanation:

Given that:

the absorption intensity = 1.00

Molarity of ethanol = 1M

NMR instrument used = 160 MHz

Temperature used = 300 K

The required concentration of ethanol can be determined as follows:

=  ( 1 \ M \times \dfrac{160\ MHz }{450 \ MHz}) \times \dfrac{300 \ K}{2.2\ K}

=  ( 1 \ M \times 0.3555 ) \times136.36}

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5 0
3 years ago
According to the weather map seen here, we could expect weather conditions in Memphis to include
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6 0
3 years ago
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To what final concentration of NH3 must a solution be adjusted to just dissolve 0.060 mol of NiC2O4 (Ksp = 4×10−10) in 1.0 L of
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Given:
0.060 mol of NiC2O4
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1.0 L of solution
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<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>

Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48

K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span> 
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M) 
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NH3 = 0.44 M

3 0
3 years ago
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