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lilavasa [31]
3 years ago
14

What was Ernest Rutherford's experiment?​

Chemistry
1 answer:
chubhunter [2.5K]3 years ago
5 0
Answer is A . He shot tiny alpha particles thought a piece of gold foil
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Determine which of the following pairs of reactants will result in a spontaneous reaction at 25°C. None of the above pairs will
Alla [95]

Answer:

Only Fe^{3+}+Mg gives spontaneous reaction.

Explanation:

A redox reaction will be spontaneous if standard reduction potential (E^{0}) of the reaction is positive. Because it leads to negative standard gibbs free energy change (\Delta G^{0}), which is a thermodynamic condition for spontaneity of a reaction.

E^{0}=E^{0}(reduction)-E^{0}(oxidation)

Where E^{0}(reduction) and E^{0}(oxidation) represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.

(1) Oxidation:         Mg-2e^{-}\rightarrow Mg^{2+} ;  E_{Mg^{2+}\mid Mg}^{0}=-2.38V

Reduction:         Fe^{3+}+3e^{-}\rightarrow Fe ; E_{Fe^{3+}\mid Fe}^{0}=-0.04V

So, E^{0}=E_{Fe^{3+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}=(-0.04+2.38)V=2.34V

Hence this pair will give spontaneous reaction.

(2) Similarly as above, E^{0}=E_{Pb^{2+}\mid Pb}^{0}-E_{Au^{+}\mid Au}^{0}=(-0.13-1.69)V=-1.82 V

Hence this pair will give non-spontaneous reaction.

(3) Similarly as above, E^{0}=E_{Ag^{+}\mid Ag}^{0}-E_{Br_{2}\mid Br^{-}}^{0}=(0.80-1.07)V=-0.27 V

Hence this pair will give non-spontaneous reaction.

(4)  Similarly as above, E^{0}=E_{Li^{+}\mid Li}^{0}-E_{Cr^{3+}\mid Cr}^{0}=(-3.04+0.74)V=-2.30 V

Hence this pair will give non-spontaneous reaction.

6 0
3 years ago
A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro
Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)

So, OH^{-} is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of OH^{-} increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant.

Therefore excess amount of OH^{-} combines with NH_{4}^{+} to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0
3 years ago
Which of the statements listed below is negative aspect of a volcanic eruption?
mixas84 [53]

Answer:

<em>C</em><em> </em><em>.</em><em> </em><em>the</em><em> </em><em>dramatic</em><em> </em><em>scenery</em><em> </em><em>created</em><em> </em><em>by</em><em> </em><em>volcanic</em><em> </em><em>eruptions</em><em> </em><em>attracts</em><em> </em><em>tourists</em><em>. </em>

5 0
3 years ago
8
love history [14]

Answer : The molar concentration of sucrose in the tea is, 0.0549 M

Explanation : Given,

Mass of sucrose = 3.765 g

Volume of solution = 0.200 L

Molar mass of sucrose  = 342.3 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of sucrose}}{\text{Molar mass of sucrose}\times \text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{3.765g}{342.3g/mole\times 0.200L}=0.0549mole/L=0.0549M

Therefore, the molar concentration of sucrose in the tea is, 0.0549 M

7 0
3 years ago
Suppose 180 ml of 3.52x10^-4 M NaOH is mixed with 220 mL of 2.47x10^-4 M MgCl2.
velikii [3]
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:

MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-

The initial concentrations are as follows:

[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
6 0
3 years ago
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