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MatroZZZ [7]
2 years ago
5

How many molecules are in the substance formula of 2C6H1206 (use coefficients)

Chemistry
1 answer:
FrozenT [24]2 years ago
6 0

<u>Answer:</u> The number of formula units present in 2 moles of C_6H_{12}O_6 are 1.2044\times 10^{24}

<u>Explanation:</u>

Formula units are defined as the number of molecules or atoms present in 1 mole of a compound or element respectively.

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of formula units

Here, 2 represents the number of moles of C_6H_{12}O_6

We are given:

Moles of C_6H_{12}O_6 (glucose) = 2 moles

Number of formula units of C_6H_{12}O_6=(2\times 6.022\times 10^{23})=1.2044\times 10^{24}

Hence, the number of formula units present in 2 moles of C_6H_{12}O_6 are 1.2044\times 10^{24}

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2 years ago
You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar
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1.04 ⨯ 10^{-4} M

<h3>Explanation</h3>

<em>A</em> = <em>ε</em> \cdot l \cdot c by the Beer-Lambert law, where

  • <em>A</em> the absorbance,
  • l the path length,
  • <em>ε</em> the molar absorptivity of the solute, and
  • c concentration of the solution.

<em>A</em> and <em>ε </em>are the same for both solutions. Therefore, l \cdot c is constant; l is inversely proportional to c. The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ 10^{-4} M. Diluting it to 45.0 mL results in a concentration of 5.17 \times 10^{-4} \times \frac{13.0}{45.0} = 1.494 M.

c is inversely related to l for the two solutions. As a result, c₂ = c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} = 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of 3.126 \times \frac{100.0}{30.0} = 1.04 ⨯ 10^{-4} M.

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