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saveliy_v [14]
3 years ago
11

Which of the following is an example of oxidation? A. Charcoal is placed on a grill and lighting fluid poured over it. B. A car

gets rusty over the course of a few years. C. Sodium bicarbonate is added to vinegar in water, and gas bubbles fizz to the surface. D. A cube of sugar dropped into a cup of hot water slowly disappears.
Chemistry
1 answer:
Vikentia [17]3 years ago
4 0

Answer:

<u>The Answer is (B)  A car get rusty over the course of few years</u>Explanation:

<u>Explanation:</u>

  • <u>Oxidation </u>refers to the process of loss of electrons by a molecule,atom or ion during a chemical reaction.The process which is just the opposite of oxidation is reduction,it occurs when their is gain of electrons .
  • <u>When iron reacts with the oxygen it forms a chemical know as Rust.In this case iron has lost some electrons and the oxygen has gained some electrons</u>

<u></u>

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What is the molarity of an aqueous solution that contains 78g of C6H12O6 dissolved in 2500 mL of solution?
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Answer:

\boxed {\boxed {\sf molarity = 0.17 \ M \ C_6H_12O_6}}

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

molarity = \frac {moles}{liters}

We are given grams of a compound and milliliters of solution, so we must make 2 conversions.

1. Gram to Moles

We must use the molar mass. First, use the Periodic Table to find the molar masses of the individual elements.

  • C: 12.011 g/mol
  • H: 1.008 g/mol
  • O: 15.999 g/mol

Next, look at the formula and note the subscripts. This tells us the number of atoms in 1 molecule. We multiply the molar mass of each element by its subscript.

6(12.011)+12(1.008)+6(15.999)=180.156 g/mol

Use this number as a ratio.

\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}

Multiply by the given number of grams.

78 \ g \ C_6H_12O_6 *\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}

Flip the fraction and divide.

78 \ g \ C_6H_12O_6 *\frac { 1 \ mol \ C_6H_12O_6}{180.156 \ g\ C_6H_12 O_6}

\frac { 78 \ mol \ C_6H_12O_6}{180.156 }= 0.432958102977 \ mol \ C_6H_12O_6

2. Milliliters to Liters

There are 1000 milliliters in 1 liter.

\frac {1 \ L }{ 1000 \ mL}

Multiply by 2500 mL.

2500 \ mL* \frac {1 \ L }{ 1000 \ mL}

2500 * \frac {1 \ L }{ 1000 }= 2.5 \ L

3. Calculate Molarity

Finally, divide the moles by the liters.

molarity = \frac {0.432958102977 \ mol \ C_6H_12O_6}{ 2.5 \ L}

molarity = 0.173183241191 \ mol \ C_6H_12O_6/L

The original measurement has 2 significant figures, so our answer must have the same. That is the hundredth place and the 3 tells us to leave the 7.

molarity \approx 0.17 \ mol \ C_6H_12O_6 /L

1 mole per liter is also equal to 1 M.

molarity = 0.17 \ M \ C_6H_12O_6

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