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2 years ago

Hello people~

Chemistry

2 answers:

Paladinen [302]2 years ago

8 0

Answer:

Polyester

Explanation:

Polyethylene terephthalate or shortly known as PET
its a clean and strong polyester
It is used in packaging

Dimas [21]2 years ago

5 0

Answer:

The answer is letter D

Have a good day

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What is the molality of a solution in which 0.42 moles aluminum chloride has been dissolved in 4200 water ?

madreJ [45]

Answer:

0.1 M

<h3>Explanation:</h3>

Molarity refers to the concentration of a solution in moles per liter.
It is calculated by dividing the number of moles of solute by the volume of solvent;

Molarity = Moles of the solute ÷ Volume of the solvent

In this case, we are given;

Number of moles of the solute, NH₄Cl as 0.42 moles
Volume of the solvent, water as 4200 mL or 4.2 L

Therefore;

Molarity = 0.42 moles ÷ 4.2 L

= 0.1 mol/L or 0.1 M

Thus, the molarity of the solution will be 0.1 M

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3 years ago

Which best describes the trends in electronegativity on the periodic table?

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Answer:

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Explanation:

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3 years ago

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

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3 years ago

Write a balanced net ionic equation for the following reaction: SrCl2(aq) + H2CO3(aq) → SrCO3(s) + HCl (aq)

max2010maxim [7]

We will balance the equation in the following order: metals, amethals, carbon, hydrogen and oxygen (the most common order).
The metal present in the equation is Sr, which is already balanced (there are 1 on each side of the equation).
The amethal present in the equation is Cl. There is 2 Cl in the left side and only one in the right side. So, we will multiply the quantity of the molecule that contains Cl by 2. Doing this, we'll obtain:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

Looking at the equation, we can see that it is now fully balanced. Hence, a balanced equation of the reaction is:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

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3 years ago

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