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3 years ago

Gamma radiation is electromagnetic energy True or false

2 answers:

8 0

True! examples of electromagnetic energy are radio waves, microwaves, infrared, visible light, ultraviolet, x-ray and gamma rays!

AfilCa [17]3 years ago

5 0

Hey there! :D

This is a true statement. Gamma radiation comes from electromagnetic energy from radioactive decay. This decay has the shortest electromagnetic wave lengths and therefore has the highest photon energy. It is extremely dangerous. Radiation in general is something to be cautious of!

I hope this helps!

~kaikers

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Need help on stoichiometry asap please

lbvjy [14]

1. Convert 135g Fe2O3 to moles, divide by 2, and multiply the resulting number by the molar mass of Al

8 0

3 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

3 years ago

Consider the reaction:

murzikaleks [220]

Answer: The formation of given amount of oxygen gas results in the absorption of 713 kJ of heat.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of oxygen gas = 83 g

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$\text{Moles of oxygen gas}=\frac{83g}{32g/mol}=2.594mol$

For the given chemical equation:

$2Fe_2O_3\rightarrow 4Fe+3O_2;\Delta H^o_{rxn}=+824.2kJ$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

By Stoichiometry of the reaction:

When 3 moles of oxygen gas is formed, the amount of heat absorbed is 824.2 kJ

So, when 2.594 moles of oxygen gas is formed, the amount of heat absorbed will be = $\frac{824.2kJ}{3mol}\times 2.59mol=713kJ$

Hence, the formation of given amount of oxygen gas results in the absorption of 713 kJ of heat.

7 0

3 years ago

Help please! please answer with a good answer

KonstantinChe [14]

Answer:

what the heck what is that?

Explanation:

5 0

3 years ago

15. How are molecules and moles different from mass?

Ierofanga [76]

Answer:

mass gives the mass of atoms while molecular weight gives the mass of molecules

7 0

3 years ago

Read 2 more answers