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8090 [49]
3 years ago
14

: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne

t ionic equation for the reaction that occurs when NaOH is added to a buffer containing H2C6H5O7 - and HC6H5O7 2- ?
Chemistry
2 answers:
olchik [2.2K]3 years ago
8 0

Answer:

The Net reaction is    

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

Explanation:

From the Question we are told that the buffers are

                H_2C_6H_5O_7^ - and  HC_6H_5O_7 ^{ 2-}

When NaOH is added the Net ionic reaction would be

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

             

Mariana [72]3 years ago
7 0

Answer:

Net ionic equation are as follows

(1) H_3C_6H_5O_7 + 0H^- \rightarrow H_2C_6H_5O_7^- +H_2O \\\\ (2)  H_2C_6H_5O_7^ - + OH^- \rightarrow H_C_6H_5O_7^{ 2-} + H_2O \\\\  (3)  HC_6H_5O_7 ^{2-} + OH^- \rightarrow C_6H_5O_7^{3-} +H_2O

Explanation:

Citric acid, H3C6H5O7, is a triprotic acid

he net ionic equation for the reaction that occurs when NaOH is added to a buffer containing H2C6H5O7 - and HC6H5O7 2-

Net ionic equation are as follows

(1) H_3C_6H_5O_7 + 0H^- \rightarrow H_2C_6H_5O_7^- +H_2O \\\\ (2)  H_2C_6H_5O_7^ - + OH^- \rightarrow H_C_6H_5O_7^{ 2-} + H_2O \\\\  (3)  HC_6H_5O_7 ^{2-} + OH^- \rightarrow C_6H_5O_7^{3-} +H_2O

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How many moles of sodium carbonate are contained by 57.3g of sodium carbonate
Lady_Fox [76]

Answer:

\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

  • Na:  22.9897693 g/mol
  • C: 12.011 g/mol
  • O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

  • Na₂ = 22.9897693 * 2= 45.9795386 g/mol
  • O₃ = 15.999 * 3= 47.997 g/mol
  • Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

6 0
3 years ago
PLS ANSWER ASAP
Kazeer [188]
The answer is c this is the answer
7 0
3 years ago
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Determine the mass (in grams) of 8.02 x 1024 atoms of mercury (Hg).
Virty [35]
<h3>Answer:</h3>

2670 g Hg

<h3>General Formulas and Concepts: </h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation: </h3>

<u>Step 1: Define</u>

8.02 × 10²⁴ atoms Hg

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of Hg - 200.59 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 8.02 \cdot 10^{24} \ atoms \ Hg(\frac{1 \ mol \ Hg}{6.022 \cdot 10^{23} \ atoms \ Hg})(\frac{200.59 \ g \ Hg}{1 \ mol \ Hg})
  2. Divide/Multiply:                                                                                               \displaystyle 2671.42 \ g \ Hg

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

2671.42 g Hg ≈ 2670 g Hg

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What mass of NaCl is dissolved in 150 g of water in a .050<br> msolution?
mart [117]

Answer:

0.4383 g

Explanation:

Molality is defined as the moles of the solute present in 1 kg of the solvent.

It is represented by 'm'.

Thus,  

Molality\ (m)=\frac {Moles\ of\ the\ solute}{Mass\ of\ the\ solvent\ (kg)}

Given that:

Mass of solvent, water = 150 g = 0.15 kg ( 1 g = 0.001 g )

Molality = 0.050 m

So,

0.050=\frac {Moles\ of\ the\ solute}{0.15}

Moles = 0.050\times 0.15\ mol= 0.0075\ mol

Molar mass of NaCl = 58.44 g/mol

Mass = Moles*Molar mass = 0.0075\times 58.44\ g = 0.4383 g

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How many dL in 1,000 mL
zalisa [80]

there are 10 dl in 1000 ml . hope it helps

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