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Licemer1 [7]
3 years ago
6

A reaction produces 0.755 mol of H2O. How many molecules of water are produced?

Chemistry
1 answer:
Fynjy0 [20]3 years ago
7 0

Answer:

4.55 * 10^23 molecules

Explanation:

This is a conversion problem. We know that one mole has 6.022 *10^23 particles or molecules so we can use this as a conversion factor.

0.755 mol ( 6.022*10^23 molecules/ 1 mol)

=4.55*10^23 molecules

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It would be a ehtanol plant
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A student is told to use 20.0 grams of sodium chloride to make an aqueous solution that has a concentration of 10.0 grams of sod
Butoxors [25]

Volume of the solution = 20.0 g NaCl * \frac{1 L solution}{10.0 g NaCl}

= 2 L solution x \frac{1000 mL}{1 L} =2000 mL

Volume of solute = 7.5 mL

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7 0
3 years ago
The molar absorptivity for aqueous solutions of phenol at 211 nm is 6.17x103L/mol/cm. Calculate the linear range of phenol conce
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Explanation:

The given data is as follows.

          \lambda = 211 nm,          \sum = 6.17 \times 10^{3} mol/L/cm

           l = 1 cm,             7% < Transmittance < 85%

Suppose the aqueous solution follows Lambert-Beer's law. Therefore,

                   Absorbance = -log \frac{\text{Percentage transmittance}}{100}

Hence, for 7% transmittance the value of absorbance will be as follows.

                  Absorbance = -log \frac{7}{100}

                           A_{1} = 1.155

For 85% transmittance the value of absorbance will be as follows.

                 Absorbance = -log \frac{85}{100}

                           A_{2} = 0.07058

According to Lambert-Beer's law.

                  A = \sum \times l \times C

where,       A = absorbance

                 \sum = molar extinction coefficient

                 C = concentration

Therefore, concentration for 7% absorbance is as follows.

                    A_{1} = \sum \times l \times C_{1}

                  C_{1} = \frac{1.155}{6.17 \times 10^{3} \times 1}

                                  = 0.187 \times 10^{-3} mol/L

                                  = 0.187 mmol/L

Concentration for 85% absorbance is as follows.

                   A_{2} = \sum \times l \times C_{2}

                  C_{2} = \frac{0.07058}{6.17 \times 10^{3} \times 1}

                                  = 0.01144 \times 10^{-3} mol/L

                                  = 0.01144 mmol/L

Thus, we can conclude that linear range of phenol concentration is 0.01144 mmol/L to 0.187 mmol/L.

6 0
3 years ago
Please answer this question ASAP. This is on a homework assignment that is due tonight!! Thank you in advance! A voltaic cell em
worty [1.4K]
Ecell = E°cell - RT/vF * lnQ

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F is the Faradays constant: 96485 J/Vmol (not sure about the mol)
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Standard conditions is 1 mol, at 298.15K and 1 atm

To find E°cell, you have to look up the reduction potensials of Fe3+ and Mg2+, and solve like this:

E°cell = cathode - anode

Cathode is where the reduction happens, so that would be the element that recieves electrons. Anode is where the oxidation happens, so that would be the element that donates electrons. In your example Fe3+ recieves electrons, and should be considered as cathode in the equation above. 

When you have found E°cell, you can just solve with the numbers I gave you. 
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