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Aleonysh [2.5K]
3 years ago
10

Nylon is a synthetic material that has many uses. It is used in the production of machine parts, cookware, and fabric. Nylon is

made from petroleum and natural gas. Suppose that there were a shortage of petroleum and natural gas. What would companies that produce nylon likely need to do? A. identify a different natural resource that nylon could be made from B. search for nylon in many different parts of the earth C. convert existing nylon back into petroleum and natural gas D. find a way to make nylon without using natural resources
Chemistry
2 answers:
ch4aika [34]3 years ago
5 0

In case of a shortage in petroleum and natural gas, a company should find a way to make nylon without using natural resources.

Nylon is a very important material used in numerous applications. As a result of this, the demand for nylon as a raw material has genuinely been on the high side ever since.

However, nylon production depends on petroleum and natural gas. It is possible that there could be a shortage of these resources since they are obtained from the environment. In such cases, a company should  find a way to make nylon without using natural resources.

Learn more about polymers: brainly.com/question/17555341

MariettaO [177]3 years ago
3 0

Answer:

identify a different natural resource that nylon could be made from

Explanation:

I just did this question

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The volume of a gas is 1.15 L at STP. At what temperature would the volume of the gas increase to 1.56 L, if the pressure is dec
denis23 [38]

A gas occupies 1.15 L at standard pressure and temperature and 1.56 L at 317 K and 650 mmHg, assuming ideal behavior.

<h3>What is an ideal gas?</h3>

An ideal gas is a gas whose behavior can be explained through ideal gas laws. One of them is the combined gas law.

A gas occupies 1.15 L (V₁) at STP (T₁ = 273,15 K and P₁ = 760 mmHg). We can calculate the temperature (T₂) at which V₂ = 1.56 L and P₂ = 650 mmHg, using the combined gas law.

\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}\\T_2 = \frac{P_2 \times V_2 \times T_1}{P_1 \times V_1} = \frac{650mmHg \times 1.56 L \times 273.15K}{760mmHg \times 1.15 L} = 317 K

A gas occupies 1.15 L at standard pressure and temperature and 1.56 L at 317 K and 650 mmHg, assuming ideal behavior.

Learn more about ideal gases here: brainly.com/question/15634266

#SPJ1

7 0
2 years ago
After mixing the solutions in a separatory funnel, the stopper should be ______ and the liquid should be _______ and the layers
maw [93]

Answer:

Hence,  

1) removed  

2) drained through the stopcock  

3) get eye level with  

4) slow the draining  

5) switch to a new flask

Explanation:

After mixing the solutions in a separatory funnel, the stopper should be removed and the liquid should be drained through the stopcock, and the layers allowed to separate. When you get close to the interface between the layers, get eye level with the funnel and turn over to slow the draining heat up until the first layer is collected. Switch to a new flask get eye level with it to collect the second layer.

7 0
3 years ago
How many grams of fosforic acid (H3PO4) are required to prepare 500 mL of a 0.2 M solution?
navik [9.2K]

Answer:

Mass = 9.8 g

Explanation:

Given data:

Molarity of solution = 0.2 M

Volume of solution = 500 mL

Number of grams of phosphoric acid = ?

Solution:

First of all we will convert the volume milliliter to litter.

500 mL × 1 L/1000 mL

0.5 L

Molarity = moles of solute / volume in litter

0.2 M =  number of moles / volume in litter

Number of moles = 0.2 mol/L × 0.5 L

Number of moles = 0.1 mol

Number of grams:

Number of moles = mass/molar mass

Mass = number of moles × molar mass

Mass = 0.1 mol × 98 g/mol

Mass = 9.8 g

7 0
3 years ago
Determine the molarity and mole fraction of a 1.09 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of a
Juli2301 [7.4K]

Answer:

Molarity = 0.809 M

mole fraction = 0.047

Explanation:

The complete question is

Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

Solution -

Solution for molarity:

1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.

1)  

Mass of 1.09 mole of acetone

= 1.09  mol x 58.0794 g/mol = 63.306 g

Density of acetone = 0.788 g/cm3  

Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3

For ethanol

1000 g divided by 0.789 g/cm3 = 1267.427 cm3

Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3  = 1.347 L

a) Molarity:

1.09 mol / 1.347 L = 0.809 M

Mole Fraction  

a) moles of ethanol:

1000 g / 46.0684 g/mol = 21.71 mol

b) moles of acetone:

1.09 / (1.09 + 21.71) = 0.047

3 0
3 years ago
Methanol, CH3OH, is a useful fuel that can be made as follows: CO(g) + 2H2(g) → CH3OH(l) A reaction mixture used 12.0 g of H2 an
algol [13]

Answer:

A = Theoretical yield = 82.24 g

B = Amount of excess reactant left = 1.72 g

Explanation:

Given data:

Mass of H₂ = 12 g

Mass of CO = 74.5 g

Theoretical yield of CH₃OH = ?

Amount of excess reactant left = ?

Solution:

First of all we will write the balance chemical equation:

CO + 2H₂  →   CH₃OH

Number of moles of H₂ = mass / molar mass

Number of moles of H₂ = 12 g/ 2 g/mol = 6 mol

Number of moles of CO = mass / molar mass

Number of moles of CO = 74.5 g/ 29 g/mol = 2.57 mol

Now we compare the moles of CH₃OH  with CO and H₂ from balance chemical equation:

                                              CO         :      CH₃OH  

                                                 1           :         1

                                                 2.57      :        2.57

   

                                                  H₂        :      CH₃OH

                                                   2          :            1

                                                    6         :          1/2 × 6 = 3 mol

The number of moles of CH₃OH produce by  CO are less so it will limiting reactant.

mass of CH₃OH  = number of moles × molar mass

mass of CH₃OH  =   2.57 mol ×   32 g/mol

mass of CH₃OH  =    82.24 g

Excess amount of H₂:

                                     CO         :         H₂

                                      1            :            2

                                  2.57         :             2×2.57 = 5.14 mol

The moles of H₂ that react with CO are 5.14. While the total number of moles of H₂ available are 6 moles. So,

The number of moles of H₂ remain untreated = 6 mol - 5.14 mol = 0.86 mol

Mass of H₂ remain untreated =  0.86 mol × 2 g/mol

Mass of H₂ remain untreated =  1.72 g

6 0
3 years ago
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