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3 years ago

What are some examples of monomers and oligomers?

1 answer:

7 0

What are some examples of monomers and oligomers?
Organic molecules, such as proteins, carbohydrates, lipids and nucleic acids, are made of simple subunits called monomers. Plasticizers are oligomeric esters widely used to soften thermoplastics such as PVC and urethane acrylate .

If a chemical compound accelerates and regulates metabolic reactions, which type of role does it play - structural or physiological?
I believe the function that it plays would be physiological since it focuses more on the regulation of the reactions inside the body.

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A 250 cm^3 solution containing 1,46 g of sodium chloride is added to an excess of silver nitrate solution. The reaction is given

faust18 [17]

Answer:

The mass of the precipitate that AgCl is 3.5803 g.

Explanation:

a) To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (NaCl) = 1.46 g

Molar mass of sulfuric acid = 58.5 g/mol

Volume of solution = $250 cm^3 =250 mL$

$1 cm^3= 1 ml$

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{1.46g\times 1000}{58.5g/mol\times 250}\\\\\text{Molarity of solution}=0.09982 M$

0.09982 M is the concentration of the sodium chloride solution.

b) $NaCl (aq)+AgNO_3 (aq)\rightarrow AgCI(s)+NaNO_3(aq)$

Moles of NaCl = $\frac{1.46 g}{58.5 g/mol}=0.02495 mol$

according to reaction 1 mol of NaCl gives 1 mol of AgCl.

Then 0.02495 moles of NaCl will give:

$\frac{1}{1}\times 0.02495 mol=0.02495 mol$ of AgCl

Mass of 0.02495 moles of AgCl:

$0.02495 mol\times 143.5 g/mol=3.5803 g$

The mass of the precipitate that AgCl is 3.5803 g.

3 0

3 years ago

True or False. In a parallel circuit, the voltage is the same anywhere in the circuit.

Harman [31]

Explanation:

1st question answer true

2nd question low resistance

hope it helps

8 0

2 years ago

Read 2 more answers

_AI + _HCI —> _H2 + _AlCl3

SOVA2 [1]

Hey there!

Al + HCl → H₂ + AlCl₃

Balance Cl.

1 on the left, 3 on the right. Add a coefficient of 3 in front of HCl.

Al + 3HCl → H₂ + AlCl₃

Balance H.

3 on the left, 2 on the right. We have to start by multiplying everything else by 2.

2Al + 3HCl → 2H₂ + 2AlCl₃

Now we have 2 on the right and 4 on the left. Change the coefficient in front of HCl from 3 to 4.

2Al + 4HCl → 2H₂ + 2AlCl₃

Now, for Cl, we have 4 on the left and 6 on the right. Change the coefficient in front of HCl again from 4 to 6.

2Al + 6HCl → 2H₂ + 2AlCl₃

Now, our H is unbalanced again. 6 on the left, 4 on the right. Change the coefficient in front of H₂ from 2 to 3.

2Al + 6HCl → 3H₂ + 2AlCl₃

Balance Al.

2 on the left, 2 on the right. Already balanced.

Here is our final balanced equation:

2Al + 6HCl → 3H₂ + 2AlCl₃

Hope this helps!

7 0

3 years ago

Grace wanted to find out the best conditions for growing lettuce plants.

Mariana [72]

Explanation:

the investigation lasts for 7 days !

hope this helps you.

8 0

2 years ago

How many liters of a 3 M NaOH stock solution would you need to make 552 mL of a 105 mM NaOH dilution

Ganezh [65]

Answer:

0.01932 L

Explanation:

First we convert 105 mM to M:

105 mM / 1000 = 0.105 M

Next we convert 552 mL to L:

552 mL / 1000 = 0.552 L

Then we use the following equation:

C₁V₁=C₂V₂

Where:

C₁ = 3 M

V₁ = ?

C₂ = 0.105 M

V₂ = 0.552 L

We input the given data:

3 M * V₁ = 0.105 M * 0.552 L

And solve for V₁:

V₁ = 0.01932 L

7 0

2 years ago