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3 years ago

In Niel’s Bohr’s model of the atom, electron move like?

Chemistry

2 answers:

iren [92.7K]3 years ago

8 0

Answer:

Pretty sure the electrons orbit.

Explanation:

Gala2k [10]3 years ago

3 0

Answer:

They move in fixed orbits around the nucleus.

Explanation:

In the Bohr model, it shows the electrons moving in fixed paths around the nucleus, making it easier to see and count them.

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Calculate the pH of a solution prepared by dissolving 0.147 moles of acetic acid and 0.405 moles of sodium acetate in water suff

Roman55 [17]

Answer:

pH = 5.19

Explanation:

Mixture of acetic acid with sodium acetate produce a buffer (Buffer is defined as the mixture of a weak acid with its conjugate base and vice versa). The pH of the buffer can be determined using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

Where pKa is -log Ka = 4.75; [A⁻] is the concentration of conjugate base (Acetate ion) and [HA] is molar concentration of the weak acid.

-You can use moles of the compounds rather than its concentrations, that is:

[HA] = 0.147 moles

[A⁻] = 0.405 moles

Replacing in H-H equation:

pH = 4.75 + log₁₀ [0.405] / [0.147]

7 0

4 years ago

A hydrated sample of cobalt(II) chloride was analyzed and found to consist of 45.44% water. What is the formula of the hydrate

valina [46]

Answer: the formula of the hydrate is CoCl2.6H2O

Explanation:Please see attachment for explanation

7 0

3 years ago

Decane (C10H22) used to produce poly(ethene). what conditions are needed?

Strike441 [17]

Cracking of hydrocarbons involves thermal decomposition.
This means that large hydrocarbon molecules break into
smaller molecules when they are heated. The hydrocarbons
are boiled and the hydrocarbon gases are either
mixed with steam and heated to a very high temperature or
passed over a hot powdered aluminium oxide catalyst.
The catalyst works by providing the hydrocarbon gases
with a convenient surface for the cracking to take place.

For example, decane (an alkane with 10 carbons)
can be cracked to produce octane and ethene.

decane octane + ethene.
C10H22(g) C8H18(g) + C2H4(g)

Octane is used as petrol.
Ethene is used in the manufacture of polymers.

Cracking an alkane produces a smaller alkane plus an alkene.
If you add up the number of hydrogen atoms in the
above reaction, you will see that there are 22 on each side.
An alkene is produced because the original alkane does
not have enough hydrogen atoms to produce two more alkanes.

hope this helps

6 0

3 years ago

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M acetic acid (Ka = 1.75x10-5) with 0.150

mylen [45]

Answer:

a) pH = 2.793

b) pH = 4.280

c) pH = 4.933

d) pH = 8.816

e) pH = 8.861

f) pH = 8.891

Explanation:

a) VNaOH = 0 mL

∴ CH3COOH ↔ CHECOO- + H3O+

⇒ Ka = 1.75 E-5 = [ H3O+ ] * [ CH3COO-] / [ CH3COOH ]

mass balance:

⇒ C CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M

charge balance:

⇒ [ H3O+ ] = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 M - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.75 E-5 [ H3O+ ] - 2.625 E-6 = 0

⇒ [ H3O+ ] = 1.61146 E-3 M

⇒ pH = - Log [ H3O+ ] = 2.793

b) after 5.0 mL NaOH:

∴ CH3COOH + NaOH ↔ CH3COONa + H2O

⇒ C NaOH = (5 E-3 L * 0.150 mol/L) / (0.025+0.01 ) = 0.02143 M

⇒ C CH3COOH = ((0.025*0.150) - (0.01*0.150)) / (0.025 + 0.01) = 0.0643 M

mass balance:

⇒ 0.02143 + 0.0643 = [ CH3COOH ] + [ CH3COO- ] = 0.086 M

charge balance:

⇒ [ H3O+ ] + [Na ] = [ CH3COO- ]

⇒ [ H3O+ ] + 0.02143 = [ CH3COO- ]

⇒ Ka = [ H3O+ ] * ( [ H3O+ ] + 0.150 ) / (0.086 - 0.02143 - [ H3O+ ]) = 1.75 E-5

⇒ [ H3O+ ]² + 0.02143 [ H3O+ ] = 1.13 E-6 - 1.75 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.02144 [ H3O+ ] - 1.13 E-6 = 0

⇒ [ H3O+ ] = 5.26 E-5 M

⇒ pH = 4.28

c) after 15 mL NaOH:

⇒ C CH3COOH = 0.0375 M

⇒ C NaOH = 0.05625 M

mass balance:

⇒ 0.09375 M = [ CH3COO- ] +[ CH3COOH ]

charge balance:

⇒ [ H3O+ ] + 0.05625 = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ] * ([ H3O+ ] + 0.05625) / (0.09375 - 0.05625 - [H3O+])

⇒ [H3O+]² + 0.05625[H3O+] = 6.5625 E-7 - 1.75 E-5 [H3O+]

⇒ [ H3O+]² + 0.05626[H3O+] - 6.5625 E-7 = 0

⇒ [ H3O+ ] = 1.1662 E-5 M

⇒ pH = 4.933

d) after 25 mL NaOH:

⇒ C NaOH = 0.075 M

⇒ C CH3COOH = 0 M....equiv. point

⇒Kh = Kw/Ka = 1 E-14 / 1.75 E-8 = 5.7143 E-10 = [ OH-]² / ( 0.075 - [OH-])

⇒ [OH-]² + 5.7143 E-10[OH-] - 4.286 E-11 = 0

⇒ [ OH- ] = 6.5463 E-6 M

⇒ pOH = 5.184

⇒ pH = 8.816

e) after 40 mL NaOH:

⇒ C NaOH = 0.0923 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 5.275 E-11 = 0

⇒ [OH-] = 7.2624 E-6 M

⇒ pOH = 5.139

⇒ pH = 8.861

f) after 60 mL NaOH:

⇒ C NaOH = 0.106 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 6.05 E-11 = 0

⇒ [OH-] = 7.7782 E-6 M

⇒ pOH = 5.11

⇒ pH = 8.891

5 0

4 years ago

PLS FAST WILL GIVE BRAINLIEST!!Substance are either ________________ or _______________.

alina1380 [7]

Answer:

elements or compounds

3 0

3 years ago