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Lady_Fox [76]
3 years ago
6

Which is the balanced chemical equation representing the reaction between

Chemistry
2 answers:
Travka [436]3 years ago
5 0

Answer:

Choice A: \rm 1\; C_2H_4 O_2 + 1\;NaHCO_3 \to 1\;NaC_2H_3O_2 + 1\;H_2O + 1\;CO_2.

Explanation:

Indeed it is possible to balance this equation by the conservation of atoms in a chemical reaction. However, knowing what's actually going on in this process will likely make this problem easier to solve.

Vinegar contains acetic acid \rm C_2H_4 O_2. Acetic acid is a monoprotic acid. In other words, each \rm C_2H_4 O_2 can dissociate to produce up to one hydrogen ion \rm H^{+}. That is:

\rm C_2H_4 O_2 \rightleftharpoons {C_2H_3 O_2}^{-} + H^{+}.

Baking soda is a common name for sodium bicarbonate \rm NaHCO_3. Each formula unit of \rm NaHCO_3 contains one bicarbonate ion: \rm {HCO_3}^{-}. Each bicarbonate ion will consume one hydrogen ion to produce water and carbon dioxide:

\rm {HCO_3}^{-} + H^{+} \to H_2O + CO_2.

For this chemical equation to balance, the number of hydrogen ions that \rm C_2H_4 O_2 supplies shall be the same as the number of these ions that \rm NaHCO_3 consumes. Each unit of \rm C_2H_4 O_2 supplies one unit of hydrogen ions while each unit of \rm NaHCO_3 consumes one unit of hydrogen ions. Reacting the two at a one-to-one ratio will make sure that this reaction neither run short of hydrogen ions or produce more hydrogen ions than it need.

Hence the coefficient in front of \rm C_2H_4 O_2 and \rm NaHCO_3 shall be the same. Let their coefficients be one.

\rm 1\; C_2H_4 O_2 + 1\;NaHCO_3 \to ?\;NaC_2H_3O_2 + ?\;H_2O + ?\;CO_2.

Now, balance this equation with reference to the number of atoms:

  • One Na atom;
  • Five H atoms;
  • Five O atoms;
  • Three C atoms.

\rm 1\; C_2H_4 O_2 + 1\;NaHCO_3 \to 1\;NaC_2H_3O_2 + 1\;H_2O + 1\;CO_2.

Citrus2011 [14]3 years ago
4 0

Answer:

The correct option is OA.

C2H4O2 + NaHCO3 - NaC2H302 + H2O + CO2

Explanation:

To solve this you have to check the number of elements in both sides of the equation.

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Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.
Svetradugi [14.3K]

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%

\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%

\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

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3 years ago
Help, I need this quick. Someone please.
timofeeve [1]

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What is the number of moles in 3.0 X 10^24 atoms of Carbon
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Explanation:

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Describe what the specific heat capacity (c) tells us about a substance. *<br> Your answer
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Answer:

it tells us of the specific amount of energy required to change the state of one mole of a substance either from solid to liquid or liquid to gas and vice versa without change in temperature

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12. What is the volume of 0.07 mol of neon gas at STP?
scoundrel [369]
<h3>Answer:</h3>

2 L Ne

<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>

  • Using Dimensional Analysis
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.07 mol Ne (g)

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

STP - 22.4 L per mole

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 0.07 \ mol \ Ne(\frac{22.4 \ L \ Ne}{1 \ mol \ Ne})
  2. Multiply:                              \displaystyle 1.568 \ L \ Ne

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

1.568 L Ne ≈ 2 L Ne

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3 years ago
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