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frosja888 [35]
3 years ago
9

Which statements accurately describe Dmitri Mendeleev’s contributions to the development of the periodic table? Check all that a

pply.
Mendeleev wrote the first modern chemistry textbook.
Mendeleev produced the first orderly arrangement of known elements.
Mendeleev ordered the elements by increasing atomic number.
Mendeleev used patterns to predict undiscovered elements.
Mendeleev accounted for variations resulting from isotopes.
Chemistry
2 answers:
Varvara68 [4.7K]3 years ago
8 0

Answer:

  • <em><u>Mendeleev produced the first orderly arrangement of known elements.</u></em>

  • <em><u>Mendeleev used patterns to predict undiscovered elements.</u></em>

Explanation:

  • <u>Mendeleev produced the first orderly arrangement of known elements and used patterns to predict the undiscovered elements.</u>

        Those two statments are true.

For the time being there were some 62 known elements. Before Medeleev some schemes to order part of the elements were proposed, but Medeleev showed the relationship between the atomic mass and the properties of the elements (supports second choice). This arrangement is known as the periodic table.

More importantly, Mendeleev predicted correctly the existance and properties of unknown elements, which is his major contribution: he left blanket spaces which where gradually filled when new elements where discovered (this supports the fourth choice).

The first modern chemistry book was written by Antoine Lavoisier (this discards first option).

Mendeleev ordered the elements by increasing mass number (this discards third choice), which was corrected later by the scientist Henry Moseley, who ordered the elements by increasing atomic number (number of protons).

Isotopes were not known by Mendeleev times, so this discards the last option.

Delvig [45]3 years ago
4 0

Answer:

The answer is B and D

Explanation:

I just took the test and got it right.

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What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250
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The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons \text{H}^{+} than hydroxide ions \text{OH}^{-}. The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that x \; \text{L} of the 0.307 \text{mol} \cdot \text{dm}^{-3} sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, x is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain 0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x moles of acetic acid and 0.307 \; x moles of acetate ions.

Let \text{HAc} denotes an acetic acid molecule and \text{Ac}^{-} denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of 10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}. There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}

By definition:

\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)

The question states that

\text{K}_{a} = 1.75 \times 10^{-5}

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10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241

Thus it takes 0.0241 \; \text{L} of sodium hydroxide to produce this buffer solution.

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