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Burka [1]
3 years ago
10

Imagine two electrons: one is close to the nucleus and one is far away. Which electron would be easier to remove from the atom a

nd why?
Chemistry
1 answer:
nataly862011 [7]3 years ago
6 0
Far away from the atom because think of it like a magnet. Nucleus is positive charge while Electron is negative charge. Put opposite charges together like a magnet its going to attract and take some effort to separate them. Now put two opposite charge a foot away from each other, you should be able to move it no problem.
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Calculate the molar mass of CH4 gas at STP when 5.46Lof the gas weighs 4g??​
meriva

Answer:

About 16.42 grams

Explanation:

PV=nRT \\\\(1)(5.46L)=n(0.0821)(273) \\\\n\approx 0.2436 \\\\4g/0.2436\approx 16.42

Hope this helps!

6 0
3 years ago
in rutherford's gold foil experiment a very small number of alpha particles were deflected. what about the structure of the atom
zvonat [6]

Answer:The atom being mostly empty space. A small number of alpha particles were deflected by large angles (> 4°) as they passed through the foil. There is a concentration of positive charge in the atom. Like charges repel, so the positive alpha particles were being repelled by positive charge

8 0
3 years ago
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sasho [114]

Answer:

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6 0
2 years ago
Balance the following redox equation in acidic solution using the smallest integers possible and select the correct coefficient
romanna [79]

Answer:

Coefficient of H^{+}(aq) is more than 4

Explanation:

Oxidation: Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

  • Balance charge: Sn^{2+}(aq)-2e^{-}\rightarrow Sn^{4+}(aq)......(1)

Reduction: Cr_{2}O_{7}^{2-}(aq)\rightarrow Cr^{3+}(aq)

  • Balance Cr: Cr_{2}O_{7}^{2-}(aq)\rightarrow 2Cr^{3+}(aq)
  • Balance O and H in acidic medium: Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)
  • Balance charge: Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l).......(2)

[3\times Equation-(1)]+Equation(2) gives balanced equation:

3Sn^{2+}(aq)+Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 3Sn^{4+}(aq)+2Cr^{3+}(aq)+7H_{2}O(l)

So coefficient of H^{+}(aq) is more than 4

3 0
3 years ago
What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe
Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]

<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒  [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0
3 years ago
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