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3 years ago

5

Why cant you use pure sodium for cooking

2 answers:

Alexus [3.1K]3 years ago

6 0

Pure sodium can severely harm your body (toxic) which can lead to death

vfiekz [6]3 years ago

5 0

It can cause explosive sodium

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Which term names the underground region between where water saturates pores and where it does not? A. water zone B. water table

Pani-rosa [81]

A)Water Table :)

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3 years ago

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What is the relationship between

pychu [463]

Answer:

As community goes through multiple changes through each stage of succession, it is not in equilibrium.

Explanation:

Equilibrium in ecology refers to a state that occurs such that a small disturbance or change is counter balanced by another change so that the community is restored to its original state.

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3 years ago

Which factor would be most likely to shrink the size of an atom’s electron cloud?

Shalnov [3]

Answer:

A. Forming a positive ion!

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6 0

3 years ago

You must make 1 L of 0.2 M acetic acid (CH3COOH). All you have available is concentrated glacial acetic acid (assay value, 98%;

Sergeu [11.5K]

Answer:

The correct answer is "11.44 ml".

Explanation:

Molarity,

= 0.2 M

Density,

= 1.05 g/ml

Volume,

= 1 L

As we know,

⇒ $Molarity=\frac{No. \ of \ moles }{Volume \ of \ solution}$

or,

⇒ $No. \ of \ moles=Molarity\times Volume$

On putting the values, we get

⇒ $=0.2\times 1$

⇒ $=0.2 \ moles$

Now,

⇒ $No. \ of \ moles=\frac{Mass \ taken}{Molecular \ mass}$

or,

⇒ $Mass \ taken=No. \ of \ moles\times Molecular \ mass$

⇒ $=0.2\times 60.05$

⇒ $=12.01 \ gram$

hence,

⇒ $Density= \frac{Mass }{Volume}$

or,

⇒ $Volume=\frac{Mass}{Density}$

⇒ $=\frac{12.01}{1.05}$

⇒ $=11.44 \ ml$

7 0

3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

<u>Answer:</u> The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

<u>Explanation:</u>

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago