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ValentinkaMS [17]
2 years ago
8

A college student drives 40 miles roundtrip each weekday to attend class. His car gets 18 miles per

Chemistry
1 answer:
Gekata [30.6K]2 years ago
3 0

In four weeks of commuting to school, it is expected the student saves a total of $13.32 if the prices went down from $2.49/gallon to $2.19.

<h3>How many gallons of gasoline does the student need each day and each week?</h3>

It is known, the total distance each day is 40 miles; moreover, the car requires 1 gallon of gasoline per 18 miles. Based on this, let's calculate the number of gallons of gasoline required:

  • 40 miles / 18 miles = 2.22 gallons of gasoline

Now, if the student requires 2.22 gallons of gasoline everyday, the total number of gallons of gasoline per week is:

  • 2.22 x 5 days = 11.1 gallons of gasoline
  • 11.1 x 4= 44.4 gallons per month

<h3>How much will the student save?</h3>

Money spent with the regular gasoline price

  • 44.4 gallons of gasoline x $2.49 = $110.55

Money spend with the new price:

  • 44.4 gallons of gasoline x $2.19 = $97.23
  • $110.55-$97.23 =$13.32

Learn more about gasoline price in: brainly.com/question/13898342

#SPJ1

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Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m
olga nikolaevna [1]

Explanation:

The reaction equation for the given reaction will be as follows.

               HCl + NaOH \rightarrow NaCl + H_{2}O

Each mole of both HCl and NaOH gives one mole of water.

Also, it is given that 1 liter of NaOH and HCl solution contains 1.6 mol dm^{-3} of NaOH (HCl).

It is known that 1 cm^{3} = 0.001 liter. So, 87 cm^{3} = 0.087 liter.

Hence, number of moles of water obtained from the given reaction are as follows.

                  0.087 liter × 1.6 mol = 0.1392 moles

            No. of moles = \frac{mass}{molar mass of water}

                0.1392 moles = \frac{mass}{18 g/mol}

                       mass = 2.5056 g

Now, volume of water present before the reaction is 2 \times 0.087 liter = 0.174 liter or 0.174 Kg (as density is 1 Kg/cm^{3}) or 174 g (as 1 kg = 1000 g).

Therefore, total weight of water present = 2.5056 g + 174 g = 176.5056 g

Formula to calculate enthalpy of neutralization is as follows.

                Enthalpy of neutralization = mS \Delta T

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where,            m = mass

                       S = specific heat capacity

                   \Delta T = change in temperature

Putting the given values in the formula as follows.

       Enthalpy of neutralization = mS \Delta T

                     = 176.5056 g \times 4.18 J/K g \times (317.4 K - 298 K)              

                                                           = 14333.73 J

   or,                                                    = 14.33 kJ

Thus, we can conclude that the enthalpy of neutralization of given reaction is 14.33 kJ.

5 0
4 years ago
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