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3 years ago

What is uniform particles ?

1 answer:

8 0

This particle is a co-polymer particle with a plastic base which can easily control arbitrary thickness. It also has excellent involtage, heat and chemical resistance.

hope this helps:)sorry if it doesnt

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1) Protons have what kind of charge?

Anuta_ua [19.1K]

1) protons have a positive charge
2) electrons have a negative charge
3) neutrons have a neutral charge

7 0

3 years ago

Step 5: Measuring the Volume of Air Near 20°C (Room Temperature) temperature of gas height of the column of gas

stealth61 [152]

Answer:

1. Temp of gas 21C, 6.2

2.Temp of gas 294, Volume of gas 0.78cm3

Explanation:

Just answered on Edge

5 0

3 years ago

Read 2 more answers

A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$

Reaction at cathode (reduction) : $Ni^{2+}+2e^-\rightarrow Ni$ $E^0_{[Ni^{2+}/Ni]}=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

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

$E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M.

New concentration of $Ni^{2+}$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(c) Now we have to calculate the concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

Now put all the given values in the above equation, we get:

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

The concentration of $Ni^{2+}$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0

4 years ago

Calculate the molality of 75.0 grams of MgCl2 (molar mass=95.21 g/mol) dissolved in 500.0 g of solvent.

nordsb [41]

<u>Answer:</u> The molality of magnesium chloride is 1.58 m

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (magnesium chloride) = 75.0

$M_{solute}$ = Molar mass of solute (magnesium chloride) = 95.21 g/mol

$W_{solvent}$ = Mass of solvent = 500.0 g

Putting values in above equation, we get:

$\text{Molality of }MgCl_2=\frac{75.0\times 1000}{95.21\times 500.0}\\\\\text{Molality of }MgCl_2=1.58m$

Hence, the molality of magnesium chloride is 1.58 m

5 0

4 years ago

Read 2 more answers

The fastest man can run 100 meters in 10 seconds. What is his speed in miles per hour?

Anestetic [448]

Answer:22.37 mph

Explanation:

100 meters = 0.062137 miles

Assume constant speed over the distance.

0.062137 miles in 10 sec

0.372822 miles in 60 sec = 1 minute

22.36932 miles in 60 minutes

22.37 mph rounded to 2 decimals

8 0

3 years ago

Read 2 more answers