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vlabodo [156]
3 years ago
8

Which one(s) below right about valence electrons?

Chemistry
2 answers:
PIT_PIT [208]3 years ago
7 0
Valence electrons are in the outermost area, have the highest energy, and are all electrons of the atom. This would be A,C and D. B might be a variable, as the Alkali Metals are highly unstable.
Electrons described by E are in Ground State.
grandymaker [24]3 years ago
3 0
Answer choice A, B, and C is right about the valence electrons. For answer choice D and E, Would in it be a more of an element kind of choice. whether the element is closer to the left side of the periodic table or not determines the high and low energy.
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State which substance is undergoing oxidation and which substance is undergoing reduction in the following reaction:
n200080 [17]

Fe is undergoing oxidation while S is undergoing reduction

<h3>What is oxidation-reduction reaction?</h3>

Oxidation-reduction reaction can simply be defined as a special type of chemical reaction in which the oxidation states of substrate change. Oxidation is the

So therefore, Fe is undergoing oxidation while S is undergoing reduction

Learn more about oxidation-reduction reaction:

brainly.com/question/17109097

#SPJ1

5 0
2 years ago
HELP me, this question not even really hard actually...
zheka24 [161]
This is a hypothesis question. This is an educated guess, basically. In my opinion, I think the tails side would hold more water.
8 0
3 years ago
When of a certain molecular compound X are dissolved in of dibenzyl ether , the freezing point of the solution is measured to be
kondaur [170]

This question is incomplete, the complete question is;

When 4.28 g of a certain molecular compound X are dissolved in 60.0 g of dibenzyl ether [(C₆H₅CH₂)₂0] , the freezing point of the solution is measured to be -3.2°C . Calculate the molar mass of X.

If you need any additional information on dibenzyl ether, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to significant digit.

Answer: molar mass of solute (X) is 88.03 g/mol

Explanation:

Given that;

mass of solute = 4.28 g

mass of solvent = 60.0 g = 0.060 kg        (Dibenzyl ether)

depression constant kf = 6.17 °CKg/mol

Freezing Point of solvent T₀ = 1.80°C       (Dibenzyl ether)

freezing point of solution Tsol = -3.20°C

Now we know that

Depression in freezing point ΔTf = depression constant kf × molaity m

and (ΔTf = T₀-Tsol)

so T₀ - Tsol = kf × m

we substitute

1.80 - (-3.20) = 6.17  × m

5 = 6.17 × m

m = 5 / 6.17

m = 0.8103 kg/mol

so molaity m = 0.8103 kg/mol

we know that

Molaity of solute m = (mass of solute / M.wt of solute) × ( 1 / mass of solvent in Kg)

solve for molar mass of solute

molar mass of solute =  (mass of solute / molaity) × ( 1 / mass of solvent in Kg)

now we substitute

molar mass = (4.28g / 0.8103 kg/mol) × (1 / 0.060kg)

molar mass = ( 5.2839 × 16.66 ) g/mol

molar mass = 88.0297 g/mol ≈ 88.03 g/mol

Therefore molar mass of solute (X) is 88.03 g/mol

3 0
3 years ago
Sodium-25 was to be used in an experiment, but it took 3.0 minutes to get the sodium from the reactor to the laboratory. If 8.0
Ksenya-84 [330]

Sodium-25 after 3 minutes : 1.0625 mg

<h3>Further explanation</h3>

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

<h3 />

No=8 mg

t1/2=60 s

T=3 min=180 s

\tt Nt=No\dfrac{1}{2}^{T/t1/2}\\\\Nt=8.5\dfrac{1}{2}^{180/60}\\\\Nt=8.5(\dfrac{1}{2})^3\\\\Nt=1.0625~mg

5 0
2 years ago
9. Circle the atom in each pair that has the greater ionization energy.
Sedbober [7]
A: BE has more ionization energy than LI

B: CA has more ionization energy than BA.
C: NA has more ionization energy than K

D: AR has more ionization energy than P

E: CI has a more ionization energy than SI
F: LI has more ionization energy than K


If any of these are wrong feel free to correct me in the comments.
7 0
2 years ago
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