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How many grams of O₂ can be prepared from the thermal decomposition of 4.27 kg of HgO? Name and calculate the mass (in kg) of th

babymother [125]

Mass of Oxygen is: 0.784gm

Potassium chlorate is KClO3 and the decomposition looks like this...

2KClO3 ==> 2KCl + 3O2

4.50 g KClO3 x 1 mole/122.55 g = 0.0367 moles

0.0367 moles KClO3 x 3 moles O2/2 moles KClO3 = 0.0245 moles O2 (see mole ratio in balanced equation)

mass of O2 = 0.0245 moles O2 x 32 g/mole = 0.784 g O2 formed

<h3>What is the pressure of a mixture of CO2 and KR?</h3>

A mixture of CO2 and Kr weighs 37.0 g and exerts a pressure of 0.737 atm in its container. is expensive, you wish to recover it from the mixture.

<h3>How much oxygen is created when one mole of potassium chlorate breaks down?</h3>

in the reaction 2K C lO3 2K C l + 3O2 K C lO3? This demonstrates that 3 moles of oxygen gas are created for every 2 moles of degraded potassium chlorate.

<h3>How is high pressure liquid CO2 produced?</h3>

High pressure liquid CO 2 is produced by compressing the gaseous CO 2 in multistage compressors to pressures in the neighbourhood of 69 bar (1000,76 psi) pressure, then cooling it to around 18 °C (64,4 °F). It is customarily filled into specially constructed steel cylinders.

Learn more about decomposition HgO:

brainly.com/question/15066243

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5 0

2 years ago

How do u do this. I'm really confused

Kaylis [27]

Are you wanting us to answer all of them I will tell you in the comment box the answer ok

8 0

4 years ago

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

An atom's tendency is to become stable. What does this mean? A. The electrons all move to one electron shell. B. The atom gains

masya89 [10]

The answer is D. To put it simply, all atom wished to become stable. The only way for that is to obtain an octet structure where the outermost shell would have 8 electrons, thus being full.

3 0

4 years ago

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Which statement best describes the revolution of a planet? A. the Earth's orbit around the sun B. the time it takes the planet t

ollegr [7]

I would d hope this helps

3 0

4 years ago

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