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3 years ago

Turning milk into plastic easy

1 answer:

4 0

You are not belive that many years ago approximate in 1900 century the people made plastic my milk this plasticity then use for making ornaments beads jewelry buckls pens and many more decorative things.

Process.

Take a thermos and add one cup of milk and heat it after heating it add 3,4 spoon of vinagar and place it aside for sometime until you will see the milk turns to plastic like shape this was very useful method for making plastic.

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What is the difference between the Lewis model and the valence-shell electron pair repulsion (VSEPR) model?

grandymaker [24]

Answer: only the VSEPR mode shows the geometric shape of a formula.

8 0

2 years ago

The mass of a hypothetical planet is 1 100 that of the earth and it’s radius is 1 4 that of earth. If a person weighs 500 n on e

avanturin [10]

Answer: Your answer is 24

Explanation:

6 0

2 years ago

Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrogen gas

Zinaida [17]

Answer: Number of molecules of hydrogen gas $6.32\times 10^{32}$

Explanation:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

Number of moles of sodium = $\frac{\text{mass of sodium}}{\text{molar mass of sodium}}=\frac{48.7 g}{23 g/mol}=2.11mol$

According to reaction , 2 moles of sodium produces 1 mole of hydrogen gas , then 2.11 mol of sodium will= $\frac{1}{2}\times 2.11 mol$ of hydrogen gas that is 1.05 moles of hydrogen gas.

Number of molecules = $N_A(\text{Avogadro number})\times$ moles of substance

Moles of hydrogen gas formed = 1.05 moles

Number of molecules of hydrogen gas = $N_A\times$ moles of hydrogen gas

Number of molecules of hydrogen gas $=6.022\times 10^{23} mole^{-1}\times 1.05 mole=6.32\times 10^{32}$

8 0

2 years ago

What evidence have u collected to explain the relationship between the Moon's revolution and lunar phrase?

Assoli18 [71]

Answer:

A lunar eclipse can only happen during a full moon.

Hope I helped :)

Explanation:

3 0

2 years ago

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

koban [17]

Answer:

1) 2.0 g

2) 0 g

3) 4.17 g

4) 2.57 g

Explanation:

First of all, we need to know the compounds and the reaction. The ion carbonate is $CO3^{-2}$ , and the ion nitrate is $NO3^{-}$ .

Sodium is in group 1, so it must lose one electron to be stable, and be the cation $Na^{+}$ . Silver has only one electron too, so the cation will be $Ag^{+}$ .

To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:

Sodium carbonate: Na2CO3

Silver nitrate: AgNO3

Silver carbonate: Ag2CO3

Sodium nitrate: NaNO3

The balanced reaction will be:

Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3

Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)

The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:

Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol

So the molar mass of the compounds must be:

Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)

AgNO3 = 170 g/mol (108 + 14 + 3x16)

Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)

NaNO3 = 85 g/mol

We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:

1 mol of Na2CO3 ---------- 2 mol of AgNO3

106 g ------------------------------ 2x170 = 340 g

x ------------------------------------ 5.14 g

By a simple direct three rule:

340x = 544.84

x = 1.6 g of Na2CO3

That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.

1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:

m = 3.6 - 1. 6 = 2.0 g

2) All the AgNO3 reacted, so there isn't a mass present after the reaction.

m = 0 g

3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3

2 moles of AgNO3 ------------- 1 mol of Ag2CO3

2x170 g ------------------------------- 276 g

5.14 g --------------------------------- x

By a simple direct three rule:

340x = 1418.64

x = 4.17 g of Ag2CO3

4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3

2 moles of AgNO3 ----------------------- 2 moles of NaNO3

2x170 g ---------------------------------------- 2x85 g

5.14 g ------------------------------------------- x

By a simple direct three rule:

340x = 873.8

x = 2.57 g

8 0

2 years ago