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Likurg_2 [28]
3 years ago
11

How many elements are in H2O

Chemistry
1 answer:
8090 [49]3 years ago
8 0

Answer: <u>b</u><u>a</u><u>l</u><u>e</u><u> </u><u>2</u><u> </u><u>e</u><u>l</u><u>e</u><u>m</u><u>e</u><u>n</u><u>t</u><u>s</u><u> </u><u>s</u><u>y</u><u>a</u><u> </u><u>a</u><u>n</u><u>g</u><u> </u><u>m</u><u>g</u><u>a</u><u> </u><u>y</u><u>o</u><u>n</u><u> </u><u>a</u><u>y</u><u> </u><u>h</u><u>y</u><u>d</u><u>r</u><u>o</u><u>g</u><u>e</u><u>n</u><u> </u><u>a</u><u>t</u><u> </u><u>o</u><u>x</u><u>y</u><u>g</u><u>e</u><u>n</u><u>.</u>

H2O is a molecule made of H for hydrogen and O for oxygen. There are 12 of these molecules. The smaller number is the number of atoms of the element to the left of it.

<em>M</em><em>A</em><em>R</em><em>K</em><em> </em><em>M</em><em>E</em><em> </em><em>A</em><em>S</em><em> </em><em>A</em><em> </em><em>B</em><em>R</em><em>A</em><em>I</em><em>N</em><em>L</em><em>I</em><em>E</em><em>S</em><em>T</em><em> </em><em>P</em><em>L</em><em>E</em><em>A</em><em>S</em><em>E</em>

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Make a drawing of the particles in an NaCl solution to show why this solution conducts electricity. Make a drawing of the partic
cluponka [151]

Answer:

  • See the image attached. It is taken from an online chemistry textbook.

  • See the explanation below.

Explanation:

<em>Sodium chloride</em> consits of sodium cations (positive ions), Na⁺, and chloride anions (negative ions), CL⁻.

<u>Pure sodium chloride</u> is packed in crystals: sodium ions and chloride ions are packed together and the ions are in fixed positions. There are not free electrons that can move. Thus, sodium chloride doesn't conduct electricity, because there are no electrons or ions which are free to move.

In aqueous solution, sodium chloride units dissociates into their ions:

         NaCl\rightarrow Na^++Cl^-

Those ions are freely to move in the solution, and such they are charge carriers, which conduct the electricity.

As explained above, in solid sodium chloride, the ions cannot move and there is not flow of current.

That is why solid pure salt of NaCl does not conduct electricity and the solutions of NaCl do conduct electricity.

The image attached show both diagrams. In the diagram A, the ions are packed together, showing that they cannot move. In the diagram B, the ions are dissolved in water, showing that they can move and carry the charge, allowing the flow of current.

5 0
3 years ago
How many mL of 6.00M HCl are needed to neutralize 50.00mL of 3.25M NaOH?
Lynna [10]

Answer:

27.00 mL

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

6 0
3 years ago
Many important biochemicals are organic acids, such as pyruvic acid ( p K a = 2.50 ) and lactic acid ( p K a = 3.86 ) . The conj
Ivenika [448]

Answer:

Pyruvic acid: conjugate base

Lactic acid: conjugate base

Explanation:

The ratio of conjugate base to conjugate acid can be found using the Henderson-Hasselbalch equation when the pH and pKa are known.

pH = pKa + log([A⁻]/[HA])

The equation can be rearranged to solve for the ratio:

pH - pKa = log([A⁻]/[HA])

[A⁻]/[HA] = 10^(pH-pKa)

Now we can calculate the ratio for the pyruvic acid:

[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 2.50) = 79433

[A⁻] = 79433[HA]

There is a much higher concentration of the conjugate base.

Similarly for lactic acid:

[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 3.86) = 3467

[A⁻] = 3467[HA]

For lactic acid the conjugate base also dominates at pH 7.4

6 0
4 years ago
A ___ is a group of atoms that remains bonded together and behaves as a single atom in a chemical reaction. Functional group rad
vladimir1956 [14]
The answer is RADICAL
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3 years ago
Read 2 more answers
Below is a proposed mechanism for the decomposition of H2O2. H2O2 + I– → H2O + IO– slow H2O2 + IO– → H2O + O2 + I– fast Which of
Savatey [412]

Answer: Option (d) is the correct answer.

Explanation:

The given equations as as follows.

       H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}           (slow)

       H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + O_{2} + I^{-}

Therefore, overall reaction equation will be as follows.

     2H_{2}O_{2} + I^{-} + IO^{-} \rightarrow 2H_{2}O + O_{2} + IO^{-} + I^{-}

So, cancelling the spectator ions then the equation will be as follows.

         H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}

As, it is known that slow step of a reaction is the rate determining step. Therefore, rate law for the slow step will be as follows.

        H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}

                 Rate law = k[H_{2}O_{2}][I^{-}]

Hence, the reaction is first order with respect to [I^{-}] and it is also first order reaction with respect to [H_{2}O_{2}].

Also, [I^{-}] acts as a catalyst in the reaction.

Thus, we can conclude that the incorrect statement is IO^{-} is a catalyst.

6 0
3 years ago
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