1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ad-work [718]
3 years ago
13

A barrel of crude oil has a volume of 42 gallons, only approximately 45% of which is processed into gasoline. If your car achiev

es 36 mi/gal, and you drive 36,000 miles in one year, how many barrels of crude oil are required to run your car for a year?
Chemistry
1 answer:
balandron [24]3 years ago
4 0

Answer:

The number of barrels of crude oil required to run the car for a year is 52.9 barrels of crude oil

Explanation:

The information given are as follows;

Percentage of crude oil processed into gasoline = 45%

Distance achieved per gallon by car = 36 mi/gal

Distance driven per year = 36,000 miles

Gasoline\,  usage \,  in \, one \,  year = \dfrac{Distance \  driven\  per \ year }{Distance\  achieved \ per gallon \ by \ car } = \dfrac{36000\, miles}{36\, miles/gallon}\dfrac{36000\, miles}{36\, miles/gallon} = 1000 \, gallons

∴ Gasoline usage in one year = 1,000 gallons

Percentage of crude oil processed into gasoline = 45%

Hence, where 45% of the crude oil produces 1000, the volume of crude, x, from which the gasoline is sourced becomes;

x × 45/100 = 1000 gallons

x =100 × 1000/45 = 20000/9 gallons of crude

1 barrel of crude oil is approximately 42 gallons of crude oil

∴ 20000/9 gallons of crude = 20000/(9×42) barells of crude = 10000/

Which gives;

52\tfrac{172}{189} \, barrels \, of \, crude \, oil

Which is 52.9 barrels of crude oil

Hence the number of barrels of crude oil required to run the car for a year = 52.9 barrels of crude oil.

You might be interested in
Which set of elements have the most similar chemical properties?
EleoNora [17]

answer: A not sure but i hope it helps :) success

7 0
3 years ago
As a quality control check, a sample of acetone is taken from a process to determine the concentration of suspended particulate
Yuri [45]

Answer:

The volume of the sample given is 850 ml, the density given is 0.79 gram per cm. Now the weight of the sample will be,  

Weight = volume × density = 850 × 0.79  

= 671.5 grams

Weight of the suspended solids given is 0.001 gram

The concentration of the sample can be determined by using the formula,  

Concentration = wt. of sample/volume

= [671.5 - 0.001) 10³ mg / 0.85 L

= 789998.82 mg/L or 789998.82 ppm

Now the concentration of suspended solids is.  

Css = 0.001 × 10³ mg / 0.85 L = 1.1764 mg per L or 1.1764 ppm

8 0
3 years ago
A solution that has a pOH of 5.36 is-
sdas [7]

Answer:

Option D

Explanation:

A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.

A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale.

The hydroxide ion molarity may be expressed as a p-function, or pOH.

pOH = −log[OH−]

Basic solutions are those with hydronium ion molarities less than 1.0 × 10−7 M and hydroxide ion molarities greater than 1.0 × 10−7 M (corresponding to pH values greater than 7.00 and pOH values less than 7.00).

7 0
3 years ago
How many particles of oxygen gas are in 22.0 grams of oxygen ?
lianna [129]

Moles of oxygen = 22g / 32 (Mr of O2)

Moles of oxygen = 0.7 moles (1 dp)

Avogadro's constant = 6.02×10^23 particles per mole

0.7 moles×(6.02×10^23) = Answer

This is assuming it wants O2, if it wants pure oxygen then replace 32 with 16.

5 0
2 years ago
A natural water with a flow of 3800 m3/d is to be treated with an alum dose of 60 mg/L. Determine the chemical feed rate for the
svet-max [94.6K]

Explanation:

First, we will calculate the feed rate of alum as follows.

   \frac{\text{60 mg alum}}{\text{1 L water}} \times \frac{\text{1000 L water}}{1 m^{3}} \times \frac{3800 m^{3}}{day} \times \frac{\text{1 g alum}}{\text{1000 mg alum}}

                  = 228000 g/day

Converting this amount into g/min as follows.

     \frac{228000 g}{1 day} \times \frac{1 day}{1440 min}

          = 158 g/min

Now, the chemical equation will be as follows.

    Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O

 \frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{3 mmol SO^{2-}_{4}}}{\text{1 mmol alum}}

      = 0.151 mmol mmol SO^{2-}_{4}/L

\frac{0.151 mmol SO^{2-}_{4}}{L} \times \frac{\text{2 meq SO^{2-}_{4}}}{\text{1 mmol SO^{2-}_{4}}} \times \frac{\text{1 meq Alk}}{\text{1 meq SO^{2-}_{4}}} \times \frac{\text{50 mg CaCO_{3}}}{\text{1 meq Alk}}

           = 15.15 mg CaCO_{3}/L

For precipitate:

Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O

  \frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{2 mmol Al(OH)_{3}}}{\text{1 mmol alum}} \times \frac{\text{78 mg Al(OH)_{3}}}{\text{1 mmol Al(OH)_{3}}}

     = 7.88 Al(OH)_{3}/L

  \frac{7.88 mg Al(OH)_{3}}{1 L} \times \frac{3800 m^{3}}{1 day} \times \frac{1000 L}{1 m^{3}} \times \frac{1 kg}{10^{6} mg}

          = 29.9 Al(OH)_{3}/day

3 0
3 years ago
Other questions:
  • How do u send pics of the answer
    13·2 answers
  • Water learned in science class that different substances release heat at different rates. He decides to test this. At home, Walt
    9·2 answers
  • How many particles are present in one mole of particles?
    6·1 answer
  • Suppose a phosphorus atom forms a bond with a fluorine atom. What type of bond must between these two elements?
    15·1 answer
  • When we think about the carbon cycle and human activities, it is important to differentiate between facts and hypotheses. Which
    12·1 answer
  • Na2CO3(aq)+2HCl(aq)→2NaCl(aq)+H2O(l)+CO2(g) A student combined two colorless aqueous solutions. One of the solutions contained N
    11·1 answer
  • A room with dimensions 7.00m×8.00m×2.50m is to be filled with pure oxygen at 22.0∘c and 1.00 atm. The molar mass of oxygen is 32
    14·1 answer
  • What material was added to powdered rock during tuttle and bowen's experiments?
    10·1 answer
  • If you have 177 grams of C4H80, how many moles do you have?
    14·1 answer
  • Pick the correct statement about the pure water. Group of answer choices Pure water contains no ions. Pure water contains equal
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!