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dolphi86 [110]
3 years ago
11

The range in size of most atomic radii is approximately _____.5 × 10−11 m to 2 × 10−10 m2 to 5 nm5 × 10−21 m to 2 × 10−20 m2 to

5 cm
Chemistry
1 answer:
Komok [63]3 years ago
4 0
A) 5*10-11 m to 2*10-10 m
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The reactants collect __solar __. ____ energy from the sun drives chemical reactions. The products include gluecose and ___oxygen____, which is released into the air

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C2H2 + 2H2 → C2H6<br> How many moles of hydrogen are needed to produce 13.78 mol of ethane?
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27.56 moles

Explanation:

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4 years ago
Which of the following reactions is a neutralization reaction? A. ZnCl2(aq) + CaCrO4(aq) → ZnCrO4(s) + CaCl2(aq) B. HNO3(aq) + L
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Neutralization reaction means the reactants must be one acid and one alkali, and the product will be H2O and metal salt.

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3 years ago
The student is now told that the four solids, in no particular order, are aluminum chloride (AlCl3), sugar (C6H12O6), benzoic ac
zaharov [31]

Answer:

AlCl₃, NaBr, C₆H₅COOH, C₆H₁₂O₆.

Explanation:

The more ions in solution, the greater the conductivity of a solution because these charged particles can carry electrons in the solution.

In the first place, when AlCl₃ dissolves in water it produces 4 ions:

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If the solution is 0.20M in AlCl₃, it will be 4 x 0.20M = 0.80M in ions.

Secondly, NaBr is a strong electrolyte (<em>complete ionization</em>) so it produces 2 moles of ions per each mole of NaBr dissolved:

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If the solution is 0.20M in NaBr, it will be 2 x 0.20M = 0.40M in ions.

Then, benzoic acid is a weak electrolyte (<em>partial ionization</em>) so it will produce <em>less than</em> 2 moles of ions per each mole of benzoic acid dissolved:

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If the solution is 0.20M in benzoic acid, the solution will be far below 0.40M in ions.

Finally, sugar has only nonpolar covalent bonds so it will produce no ions in solution, thus being a poor electricity conductor.

4 0
4 years ago
For hydrocyanic acid, HCN, Ka = 4.9 × 10-10. Calculate the pH of 0.20 M NaCN. What is the concentration of HCN in the solution?
gogolik [260]
By using the ICE table :

initial    0.2 M            0           0
change -X                 + X           +X
Equ     (0.2 -X)            X               X

when Ka = (X) (X) / (0.2-X)
so by substitution:

4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X 
∴X ≈ 10^-6
∴[HCN] = 10^-6

and PH = -㏒[H+]
             = -㏒ 10^-6
             = 6
3 0
3 years ago
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