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2 years ago

Define the following terms - you may need to consult your lecture text or other suitable resource:

Chemistry

1 answer:

natta225 [31]2 years ago

6 0

Answer:

a) Monomers: monomers are unit molecules, that can react together with other monomers, to form a long chain molecule called a polymer. Th polymer formed can also be in a three dimensional network. The process of this conversion of monomers to polymers is called polymerization.

b) Repeating unit: A repeating unit is a unit of the polymer formed, whose repetition would produce a long complete polymer chain. A polymer is made up of these repeating links of molecules that form a long chain of molecules.

c) Condensation polymerization: This is a form of condensation reaction, that involves the combination of molecules into polymers with the loss of small molecules such as water or methanol as by products.

d) Cross-linked polymer: This is a polymer formed from a type of bonding of molecules. The bonding is usually in the form of covalent bonds or ionic bonds and the polymers can be either synthetic polymers or natural polymers. The cross-links leads to an alteration in the physical properties of the polymer.

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If a pharmacist added 12 g of azelaic acid diluent should be used to prepare 8 fluidto 50 g of an ointment containing 15% ounces

natta225 [31]

Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-azelaic acid) = 12g.
mr₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
mr₃ = mr₂ + mr₂ = 50g + 12g = 62g.
ω₃ = mrs₃÷mr₃ = 19,5g ÷ 62g = 31,45% = 0,3145.

7 0

2 years ago

How many moles of CaC2 are needed to react completely with 45.0g H2O?

sertanlavr [38]

Answer:

In order to react with 45 g of water 1.25 moles of CaC₂ are required.

Explanation:

Given data:

Moles of CaC₂ needed = ?

Mass of water = 45.0 g

Solution:

Chemical equation:

CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 45 g/ 18 g/mol

Number of moles = 2.5 mol

Now we will compare the moles of water and CaC₂ from balance chemical equation:

H₂O : CaC₂

2 : 1

2.5 : 1/2×2.5 =1.25 mol

In order to react with 45 g of water 1.25 moles of CaC₂ are required.

6 0

2 years ago

Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A

lesantik [10]

Answer: The chemical equations and equilibrium constant expression for each ionization steps is written below.

Explanation:

The chemical formula of carbonic acid is $H_2CO_3$ . It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water

The chemical equation for the first dissociation of carbonic acid follows:

$H_2CO_3(aq.)\rightleftharpoons H^+(aq.)+HCO_3^-(aq.)$

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}$

The chemical equation for the second dissociation of carbonic acid follows:

$HCO_3^-(aq.)\rightarrow H^+(aq.)+CO_3^{2-}(aq.)$

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}$

Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.

6 0

2 years ago

How many moles of N2 are essential for generating 0.08 moles of Li3N in the given reaction?6Li N2 2Li3N

jeka57 [31]

08 moles Li3N * 1mole N2/2moles Li3N = 0.04

3 0

2 years ago

Read 2 more answers

I NEED HELP ASAP, WILL MARK BRAINLEST!

Andre45 [30]

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial SiCl₄ O₂ SiO₂

1 120 g 240 g 38.2 g

2 75 g 50 g 25.2 g

Percentage yield for trial 1

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

.706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

actual/theoretical x 100

38.2 / 42.44 = .900 = 90.0% yield

Leftover reactant for trial 1

We know oxygen is the excess reactant.

Mass to moles:

molar mass O₂ = 32.00 g/mol

240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

7.5 - .706 = 6.8 moles left over

Moles to mass:

6.8 mol O₂ x 32.00g/mol = 217.4 g O₂

Percentage yield for trial 2

Mass to moles:

molar mass SiCl₄: 169.9 g/mol

75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

molar mass SiO₂: 60.09 g/mol

.441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

actual/theoretical x 100

25.2 / 26.5 = .950 = 95.0% yield

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

6 0

2 years ago