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timama [110]
3 years ago
7

Copper, a metal known since ancient times, exists in two stable isotopic forms, 6329Cu (69.09%) and 6529Cu (30.91%). Their atomi

c masses are 62.93 amu and 64.9278 amu respectively. Calculate the average atomic mass of copper.
Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
7 0

Answer:

Wt. Avg. Atomic Weight => 63.35457 amu

Explanation:

Given              Isotopic             %Abundance     fractional          Wt Avg

                  At. Mass (amu)                                  abundance      contribution                    

Cu-63              62.93                      69.09              0.6909            43.4783

Cu-65              64.9278                  20.0668        0.200668        20.0668

Wt Average of all isotopes = ∑Wt Avg Contributions

=  43.4783 amu + 20.0668 amu = 63.35457 amu

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Jim wants to react hydrogen and oxygen to get 36 grams of water. If he starts with 4 grams of Hydrogen (H), then how many grams
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Answer:

Mass of Oxygen = 32 grams

Explanation:

Given:

Mass of water = 36 grams

Mass of Hydrogen = 4 grams

Find:

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Mass of water = Mass of Hydrogen + Mass of Oxygen

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Heat moves from what source by convection and radiation?
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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
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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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