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1 year ago

PLEASE HELP I do not understand

Chemistry

2 answers:

Vikentia [17]1 year ago

4 0

Answer:

Absorbing beta particle because the beta is the numbers and are less and the big numbers are positive and they are the alpha so when you add beta particle it is called Absorbing so the answer is Absorbing beta particle

amid [387]1 year ago

4 0

Answer:

absorbing beta particle

Explanation:

radioactivity general facts.

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Which of the following is NOT true about relative dating?

Igoryamba

Answer:

It uses fossils to help pinpoint the ages of rocks.

Explanation:

Radiocarbon dating can not be used to determine the age of rocks.

Carbon dating works well only for objects that are less than 50,000 years. Most rocks are far older than that. Over time, carbon-14 decays gradually into nitrogen. Hence, we can not really use radiocarbon dating to determine the absolute age of a rock sample since the carbon-14 in the fossils of ancient rock samples may have completely decayed.

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2 years ago

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2 years ago

A solution has a concetration of 0.3mol/dm3 of sodium hydroxide .what volume of

HACTEHA [7]

Answer:

Ba(OH)2 + H2SO4 ------> BaSO4 + 2H2O

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7 0

2 years ago

What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.

OLga [1]

Answer : The voltage of the voltaic cell is 1.14 V

Explanation :

From the given cell representation, we conclude that

The copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

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

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

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$E^o_{cell}=(+0.34V)-(-0.76V)$

$E^o_{cell}=1.1V$

Now we have to calculate the emf or voltage of the cell.

Using Nernest equation at 298 K :

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

where,

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

$E_{cell}$ = ?

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

$E_{cell}=1.1-\frac{0.0592}{2}\log \frac{0.2}{4.0}$

$E_{cell}=1.14V$

Therefore, the voltage of the voltaic cell is 1.14 V

5 0

2 years ago

Use the unbalanced equation NH3 + O2 = NO + H2O, what is the mole ratio between NH3 and NO

kirill115 [55]

<span>The first thing that needs to be done in order to answer the question above is to balance first the chemical equation by seeing to that the number of moles of a certain element on the reactant side is equal to the number of moles in the product side. 4NH3 + 5O2  4NO + 6H2O The mole fraction between the NH3 and NO is therefore 4:4 or 1:1.</span>

4 0

2 years ago

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