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3 years ago

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Based on the synthesis reaction, what would the product of the reaction be? NaPO3 + CuO → ? NaO + CuPO3 Na + P + Cu + O4 NaCuPO4

NaCu + PO4

2 answers:

12345 [234]3 years ago

6 0

Answer:

C

Explanation:

trapecia [35]3 years ago

5 0

Answer:

C

Explanation:

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The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x

Andrews [41]

Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /</span>λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules

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

The concentration of calcium in a cell is 0.3%. The concentration of calcium in the surrounding fluid is 0.1%. How could the cel

iragen [17]

Answer:

Primary active transport

Explanation:

In a cell, the movement of molecules like calcium ions (Ca²⁺), <em>to a region having high solute concentration from a region having low solute concentration, through the cell membrane requires metabolic energy</em> and is known as Primary active transport.

It is given that the concentration of calcium in the cell (0.3%) is greater than the concentration of calcium in the fluid surrounding the cell (0.1%). <em><u>So the calcium ions move into the cell and the cell obtains more calcium. </u></em>

<u>Therefore, the cell obtains more calcium by the process of Primary active transport.</u>

4 0

3 years ago

Which types of atomic orbitals of the central atom mix to form hybrid orbitals in:<br> (b) CS₂;

Anettt [7]

<u>One s orbital</u> and <u>one p orbital</u> are the exact types of atomic orbitals of the central atom mix to form hybrid orbitals in CS₂

<h3>What is atomic orbital?</h3>

An atomic orbital is a function used in atomic theory and quantum mechanics to explain the position and wave-like behaviour of an electron in an atom. This function can be used to determine the likelihood of discovering any atom's electron in any particular area surrounding the nucleus.

The physical area or space where the electron may be calculated to be present, as predicted by the specific mathematical shape of the orbital, is referred to as an atomic orbital.

The three quantum numbers n, l, and $m_l$ which correspond to the electron's energy, angular momentum, and an angular momentum vector component, are used to describe all orbitals in an atom (magnetic quantum number).

Learn more about Atomic Orbital

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6 0

1 year ago

When potassium chloride breaks down, potassium metal and chlorine gas are formed. What type of reaction is this?

Anni [7]

Answer:

Chemical reaction

Explanation:

The physical form of potassium metal and chlorine gas are being formed, meaning there is an involvement of the rearrangement of the molecular or ionic structure.

6 0

3 years ago