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2 years ago

Which subatomic particle can vary between NEUTRAL atoms of the SAME element

Chemistry

1 answer:

tekilochka [14]2 years ago

7 0

Answer:

answer is-neutrons

Explanation:

fj’yf’jg’kh’jy’tfthfy,gtj ftjfrvf

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What type of molecule is shown below

Kobotan [32]

Answer is: C alkane.

Name of this molecule is n-hexane.

Hexane is alkane (acyclic saturated hydrocarbon, carbon-carbon bonds are single) of six carbon atoms.

Carbon atoms in hexane have sp3 hybridization.

In sp3 hybridization hybridize one s-orbital and three p-orbitals of carbon atom.

Alkene has one double bond and alkyne has one triple bond.

4 0

3 years ago

A comic book villain is holding you at gun point and is making you drink a sample of acid he gives you a beaker with 100 ml of a

miss Akunina [59]

Answer:

b. 10 mL

Explanation:

First we calculate the amount of H⁺ moles in the acid:

pH = -log [H⁺]

[H⁺] = $10^{-pH}$

[H⁺] = 10⁻⁵ = 1x10⁻⁵M

100 mL ⇒ 100 / 1000 = 0.100 L

1x10⁻⁵M * 0.100 L = 1x10⁻⁶ mol H⁺

In order to have a neutral solution we would need the same amount of OH⁻ moles.

We can use the pOH value of the strong base:

pOH = 14 - pH

pOH = 14 - 10 = 4

Then we calculate the molar concentration of the OH⁻ species in the basic solution:

pOH = -log [OH⁻]

[OH⁻] = $10^{-pOH}$ = 1x10⁻⁴ M

If we use 10 mL of the basic solution the number of OH⁻ would be:

10 mL ⇒ 10 / 1000 = 0.010 L

1x10⁻⁴ M * 0.010 L = 1x10⁻⁶ mol OH⁻

It would be equal to the moles of H⁺ so the answer is b.

7 0

3 years ago

Read 2 more answers

How are cyanobacteria related to increases in oxygen in the atmosphere

k0ka [10]

This is a biology question. Cyanobacteria partake in photsynthesis, meaning they take in carbon dioxide and water to produce glucose and oxygen.

7 0

3 years ago

5. Wskaż liczbę protonów w atomie glinu?

Scrat [10]

Answer:

write english bro

Explanation:

pls i cant understand

3 0

3 years ago

Calculate the equilibrium concentration of c2o42− in a 0.20 m solution of oxalic acid.

spin [16.1K]

Answer:

[C2O2−4] =1.5⋅10−4⋅mol⋅dm−3

Explanation:

For the datasheet found at Chemistry Libretext,

Ka1=5.6⋅10−2 and Ka2=1.5⋅10−4 [1]

for the separation of the primary and second nucleon once oxalic corrosive C2H2O4 breaks up in water at 25oC (298⋅K).

Build the RICE table (in moles per l, mol⋅dm−3, or identically M) for the separation of the primary oxalic nucleon. offer the growth access H+(aq) fixation be x⋅mol⋅dm−3.

R C2H2O4(aq)⇌C2HO−4(aq)+H+(aq)

I 0.20

C −x +x

E 0.20−x x

By definition,

Ka1=[C2HO−4(aq)][H+(aq)][C2H2O4(aq)]=5.6⋅10−2

Improving the articulation can provides a quadratic condition regarding x, sinking for x provides [C2HO−4]=x=8.13⋅10−2⋅mol⋅dm−3

(dispose of the negative arrangement since fixations can faithfully be additional noteworthy or appreciate zero).

Presently build a second RICE table, for the separation of the second oxalic nucleon from the amphoteric C2HO−4(aq) particle. offer the modification access C2O2−4(aq) be +y⋅mol⋅dm−3. None of those species was out there within the underlying arrangement. (Kw is neglectable) therefore the underlying centralization of each C2HO−4 and H+ are going to be appreciate that at the harmony position of the first ionization response.

R C2HO−4(aq)⇌C2O2−4(aq)+H+(aq)

I 8.13⋅10−2 zero eight.13⋅10−2

C −y +y

E 8.13⋅10−2−y y eight.13⋅10−2+y

It is smart to just accept that

a. 8.13⋅10−2−y≈8.13⋅10−2,

b. 8.13⋅10−2+y≈8.13⋅10−2 , and

c. The separation of C2HO−4(aq)

Accordingly

Ka2=[C2O2−4(aq)][H+(aq)][C2HO−4(aq)]=1.5⋅10−4≈(8.13⋅10−2 )⋅y8.13⋅10−2

Consequently [C2O2−4(aq)]=y≈Ka2=1.5⋅10−4]⋅mol⋅dm−3

3 0

3 years ago