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just olya [345]
3 years ago
8

All of the following can be used to define a base except (3 points) Question 1 options: 1) a hydronium ion donor in a reaction 2

) a substance that increases the concentration of hydroxide ions 3) an electron pair donor in a reaction 4) a substance that is a hydrogen ion acceptor in a reaction
Chemistry
1 answer:
uranmaximum [27]3 years ago
4 0
Answer is: <span>1) a hydronium ion donor in a reaction.
Base is hydroxide ion (OH</span>⁻) donor, not hydronium ion (H₃O⁺).
Lewis base <span>can donate its lone pair of electrons to Lewis acid.
Base </span><span>increases the concentration of hydroxide ions, for example sodium hydroxide or barium hydroxide in water NaOH </span>→ Na⁺ + OH⁻.
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The balanced chemical equation for the production of chromium metal from the reaction of chromium(ll) nitrate reacts with a strip of zinc is:

3 Zn + 2 Cr(NO₃)₃ → 2 Cr + 3 Zn(NO₃)₂

This is a redox reaction, which <u>is a chemical reaction in which one or more electrons are transferred between the reagents</u>, causing a change in their oxidation states. In the proposed reaction, Cr oxidation state goes from +3 to 0, becoming metallic chromium, while Zn goes from being Zn⁰ to Zn²⁺.

<u>The mass of chromium metal produced in the above reaction will be,</u>

425.0 mL x \frac{1 L}{1000 mL} x  \frac{0.25 mol Cr(NO_{3})_{3}  }{1 L} x \frac{2 mol Cr  }{2 mol Cr(NO_{3})_{3} } x \frac{51.9961 g Cr}{1 mol Cr} = 5.52 g

So, the mass of chromium metal produced when 425.0mL of 0.25M chromium(ll) nitrate reacts with a strip of zinc that remains in excess is 5.52 g of Cr.

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After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of
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b) The dehydrated sample absorbed moisture after heating

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a) Strong initial heating caused some of the hydrate sample to splatter out.

This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).

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Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.

c) The amount of the hydrate sample used was too small.

It will create some errors but they do not create a difference of 13% difference as stated in the problem.

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Here the error is small.

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Answer: There are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

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According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

So, the number of atoms present in 0.67 moles are as follows.

0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms

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