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VashaNatasha [74]

3 years ago

15

These two beakers represent solutions of HCl and NaOH. Describe a third beaker showing the ions that remain after the reaction h

as gone to completion.
The third beaker contains __ OH- ion(s), __ Cl- ion(s), __ Na+ ion(s), and __ H+ ion(s)

1 answer:

DENIUS [597]3 years ago

7 0

Number of ions in the third beaker:

4 Cl⁻ ions

5 Na⁺ ions

1 OH⁻ ion

Explanation:

If you mix the backers the following chemical reaction will occur:

HCl + NaOH → NaCl + H₂O

In the ionic form you write the equation:

H⁺ (aq) + Cl⁻ (aq) + Na⁺ (aq) + OH⁻ (aq) → Cl⁻ (aq) + Na⁺ (aq) + H₂O (l)

where:

(aq) - aqueous (dissolved in water)

(l) - liquid

Number of ions in the first beaker:

4 Cl⁻ ions

4 H⁺ ions

Number of ions in the second beaker:

5 Na⁺ ions

5 OH⁻ ions

Number of ions in the third beaker:

4 Cl⁻ ions

5 Na⁺ ions

1 OH⁻ ion

The rest of H⁺ and OH⁻ ion reacted to form water.

Learn more about:

ions

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Ipatiy [6.2K]

Answer:

The molecular formula of estradiol is: $C_{18}H_{24}O_2$ .

Explanation:

Molar mass of of estradiol = M= 272.37 g/mol

Let the molecular formula of estradiol be $C_xH_yO_z$

Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100$

Percentage of carbon in estradiol :

$\frac{x\times 12 g/mol}{272.37 g/mol}\times 100=79.37\%$

x = 18.0

Percentage of hydrogen in estradiol :

$\frac{y\times 1g/mol}{272.37 g/mol}\times 100=8.88\%$

y = 24.2 ≈ 24

Percentage of oxygen in estradiol :

$\frac{z\times 16 g/mol}{272.37 g/mol}\times 100=11.75\%$

z = 2

The molecular formula of estradiol is: $C_xH_yO_z:C_{18}H_{24}O_2$

7 0

3 years ago

Hydrochloride acid +_________ zinc chloride +hydrogen

Klio2033 [76]

Answer:

Hydrochloride acid + Zinc = Zinc Chloride + Hydrogen

Explanation:

When Hydrochloride acid and Zinc react, it results in the formation of Zinc chloride and hydrogen.

<em>Hope I helped</em>

3 0

2 years ago

10) Give two example of where you would find elements in our daily lives.

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Carbon is found in oil and gas.
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6 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

In which of the following are the symbol and name for the ion given correctly?

Andrej [43]

Fe2+:ferrous ion; fe3+ ferric ion are the one which are correctly named(Answer A) Sn2+ are called stannous ion while Sn4+ are known as stannic ions. Pb2+ are called lead (II) ions while Pb4+ are known as lead (IV) ion Co2+ are known was cobalt (II) ions while Co3+ are known as cobalt(IIi) ion.

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3 years ago