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3 years ago

Compare and contrast the elements Potassium, K, and Neon, Ne. Use at least ten comparisons or descriptions.

Chemistry

1 answer:

vovikov84 [41]3 years ago

4 0

Answer:

K is a group 1 element while Ne is a group 0 element

K is a good reducing agent while Ne is inert

K is metal while Ne is a gas

K has one valence electron while Ne has no valence electron

Explanation:

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steam and methane

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3 years ago

Need help with 22 and 24

choli [55]

Answer:

22:

Formular:

$atomic \: mass = \frac{ \sum(isotopic \: mass \times \%abundance)}{100} \\$

substitute:

$atomic \: mass = \frac{(23.985 \times 78.70) + (24.986 \times 10.13) + (25.983 \times 11.17)}{100} \\ \\ = \frac{(1887.620) + (253.108) + (290.230)}{100} \\ \\ = \frac{2430.958}{100} \\ \\ { \boxed{ \boxed{average \: atomic \: mass = 24.3 \: amu}}}$

23:

Same element is represented by same number of protons. 

Answer:

6 protons. 6 protons

7 neutrons. 8 neutrons

6 electrons. 6 electrons

Note: Atoms with same proton number but different mass number are called isotopes

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3 years ago

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When 50.0 ml of a 0.3000 m agno3 solution is added to 50.0 ml of a solution of mgcl2, an agcl precipitate forms immediately. the

frosja888 [35]

0.0165 m The balanced equation for the reaction is AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2 So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights. Atomic weight silver = 107.8682 Atomic weight chlorine = 35.453 Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol Now how many moles were produced? 0.1183 g / 143.3212 g/mol = 0.000825419 mol So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division. 0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m Rounding to 3 significant figures gives 0.0165 m

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4 years ago

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What is the mass of 3.0 x 1023 atoms of neon

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3 years ago

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G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

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3 years ago