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Hitman42 [59]
3 years ago
14

Which substance is important in making plastics?

Chemistry
2 answers:
QveST [7]3 years ago
8 0
Based on the compounds given, I can say that the substance that is considered essential in the plastic making is NITRIC ACID. This substance is also known as aqua fortis and spirit of niter wherein this decomposes into oxides of nitrogen and water. The answer would be the last option. Hope this helps.
gregori [183]3 years ago
8 0
<h3><u>Answer;</u></h3>

Nitric acid

<h3><u>Explanation;</u></h3>
  • Nitric acid is a corrosive liquid inorganic acid HNO3 used mostly as an oxidizing agent, and in making  organic compounds.
  • <em><u>Nitric acid is used in the production of ammonium nitrate for fertilizers, making plastics, and in the manufacture of dyes.</u></em> It is also used for making explosives such as nitroglycerin and TNT.


You might be interested in
When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)
Firdavs [7]

Answer:

Mass of Ba₃(PO₄)₂ = 0.0361 g

Explanation:

Given data:

Volume of Ba(NO₃)₂ = 1.2 mL (1.2 × 10⁻³ L )

Molarity of Ba(NO₃)₂ = 0.152 M

Volume of K₃PO₄ = 4.2 mL (4.2 × 10⁻³ L)

Molarity of K₃PO₄ =  0.604 M

Mass of Ba₃(PO₄)₂ produced = ?

Solution:

Chemical equation:

3Ba(NO₃)₂  + 2K₃PO₄  → Ba₃(PO₄)₂  + 6KNO₃

Number of moles of Ba(NO₃)₂ = Molarity × Volume in litter

Number of moles of Ba(NO₃)₂ = 0.152 M × 1.2 × 10⁻³ L

Number of moles of Ba(NO₃)₂ = 0.182 × 10⁻³ mol

Number of moles of K₃PO₄ = Molarity × Volume in litter

Number of moles of K₃PO₄ = 0.604 M × 4.2 × 10⁻³ L

Number of moles of K₃PO₄ = 2.537 × 10⁻³ mol

Now we will compare the moles of Ba₃(PO₄)₂ with K₃PO₄ and Ba(NO₃)₂ .

              Ba(NO₃)₂        :         Ba₃(PO₄)₂

                   3                :               1

              0.182 × 10⁻³    :              1/3 ×0.182 × 10⁻³ = 0.060 × 10⁻³ mol

                K₃PO₄           :          Ba₃(PO₄)₂

                   2                 :                1

              2.537 × 10⁻³     :               1/2 ×  2.537 × 10⁻³= 1.269 × 10⁻³ mol

The number of moles of Ba₃(PO₄)₂ produced by  Ba(NO₃)₂  are less it will limiting reactant.

Mass of Ba₃(PO₄)₂ = moles × molar mass

Mass of Ba₃(PO₄)₂ = 0.060 × 10⁻³ mol × 601.93 g/mol

Mass of Ba₃(PO₄)₂ = 36.12 × 10⁻³ g

Mass of Ba₃(PO₄)₂ = 0.0361 g

6 0
3 years ago
Calculate the theoretical yield of aspirin obtained when 2.0 g of salicylic acid and 5.0 mL of acetic anhydride (density = 1.08
saw5 [17]

Answer:

The percent yield would be 73%

Explanation:

The balanced reaction for the obtention of acetylsalicylic acid (aspirin) is the following:

Salicylic acid + acetic anhydride → acetylsalicylic acid + acetic acid

C₇H₆O₃          +          C₄H₆O₃        →         C₉H₈O₄           +    C₂H₄O₂

According to the reaction, 1 mol of salicylic acid reacts with 1 mol of acetic anhydride to give 1 mol of acetylsalicylic acid (aspirin) and 1 mol of acetic acid.

1 mol aspirin (C₉H₈O₄) = (9 x 12 g/mol) + (8 x 1 g/mol) + (4 x 16 g/mol)

                                     = 180 g

1 mol salicylic acid (C₇H₆O₃) = (7 x 12 g/mol) + (6 x 1 g/mol) + (3 x 16 g/mol)

                                              = 138 g

1 mol acetic anhydride (C₄H₆O₃) = (4 x 12 g/mol) + (6 x 1 g/mol) + (3 x 16 g/mol)

                                                      = 102 g

The stoichiometric ratio is = 138 g salicylic acid/102 g acetic anhydride= 1.35

We have:

2.0 g salicylic acid

acetic anhydride = 5.0 mL x 1.08 g/mL = 5.4 g

The reactants ratio is = 2.0 g salicylic acid/5.4 g acetic anhydride = 0.37

0.37 < 1.35 , therefore <em>salicylic acid is the limiting reactant</em>.

Now, we use the amount of salicylic acid to calculate the theoretical amount of aspirin. For this, we know that 1 mol of aspirin (180 g) is obtained from 1 mol of salycilic acid (138 g):

theoretical yield= 180 g aspirin/138 g salycilic acid x 2.0 g salycilic acid = 2.61 g aspirin

actual yield = 1.9 g

Finally, we calculate the yield:

percent yield = actual yield/theoretical yield x 100

        = 1.9 g/2.6 g x 100 = 72.8% ≅ 73%

6 0
2 years ago
Someone please help me!!!!
Artyom0805 [142]

Double replacement :

2Na₃PO₄+3CaCl₂⇒6NaCl + Ca₃(PO₄)₂

<h3>Further explanation</h3>

1. A single replacement reaction is a chemical reaction in which one element replaces the other elements of a compound to produce new elements and compounds

2. Double-Replacement reactions. Happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

3. Combination/syntesis : 2 or more reactants combine to form a new compound

4. Decomposition : the reactant is decomposed into 2/more products

If we look at the reaction options available, all of them can be included in the double replacement reaction, but we only choose the reaction from Sodium phosphate and Calcium chloride which leads to options: C because it is balanced (from the number of atoms in the same reactant and product) and is in accordance with the chemical formula of each compound (both products and reactants)

3 0
3 years ago
Someone Please HELP Me
scoray [572]
Use a calculator to add those thank you ur welcome
7 0
3 years ago
Find the theoretical oxygen demand for the
never [62]

Answer:

a) 213.3 mg/L

b) 62.61 mg/L

c) 0.0225 mg/L

Explanation:

Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products

a) Given:

Concentration of acetic acid,[CH3COOH] = 200 mg/L

CH3COOH + 2O2 \rightarrow 2CO2 + 2H2O

ThOD = \frac{mass\ O2}{mass\ CH3COOH} * conc. CH3COOH

Based on the reaction stoichiometry:

mass of CH3COOH = 60 g

mass of O2= 2(32) = 64 g

ThOD = \frac{64 g}{60 g} * 200mg/L = 213.3 mg/L

b) Given:

Concentration of ethanol, [C2H5OH] = 30 mg/L

C2H5OH + 3O2 \rightarrow 2CO2 + 3H2O

ThOD = \frac{mass\ O2}{mass\ C2H5OH} * conc. C2H5OH

Based on the reaction stoichiometry:

mass of C2H5OH = 46 g

mass of O2= 3(32) = 96 g

ThOD = \frac{96 g}{46 g} * 30mg/L = 62.61 mg/L

c) Given:

Concentration of sucrose, [C12H22O11] = 50 mg/L

C12H22O11 + 12O2 \rightarrow 12CO2 + 11H2O

ThOD = \frac{mass\ O2}{mass\ C22H22O11} * conc. C12H22O11

Based on the reaction stoichiometry:

mass of C12H22O11 = 342 g

mass of O2= 12(32) = 384 g

ThOD = \frac{384 g}{342 g} * 50mg/L = 0.0225 mg/L

7 0
3 years ago
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