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3 years ago

6.0 g of copper was heated from 20 degree c to 90 degree c . How much energy was used to heat cu?

1 answer:

4 0

Copper heat capacity would be 0.385J/C*gram which means it needs 0.385 Joule of energy to increase 1 gram of copper temperature by 1 Celcius. The calculation would be:
energy= heat capacity *mass * temperature difference
energy= 0.385J/C*gram * 6g * (90-20)
energy= 161.7J

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A spinning turbine can generate electricity only in the form of a/an _______ current.

Tom [10]

Answer:

The correct answer is - alternating and direct, in order.

Explanation:

Alternating current is is type of electric current that is characterized by the direction of the flow of electrons in continuously switches its directs in opposite manner at regular cycles. While direct current or DC is flow of the electrons that move from starting to end in one direction.

Spinning turbines always leads to the alternating electric current while only solar energy produces the direct current with the help of the solar panels.

Thus, the correct answer is - alternating and direct, in order.

3 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

Why does every plant cell have a nucleus

Lena [83]

Every plant cell has a nucleus because a nucleus controls the cell's main activities.

3 0

3 years ago

Read 2 more answers

madam [21]

The answer is D. 30 percent

5 0

4 years ago

How many moles of carbon are in 7.87x10^7 carbon molecules?

zaharov [31]

There are 6.022*10^23 molecules in 1 mole of carbon
So how many will moles will be 7.87*20^7?
Let the required number of moles be ‘x’.
1 mole ———6.022*10^23
x moles———7.87*10^7
(Cross multiplication)
x=7.87*10^7/6.022*10^23
Therefore x=1.3*10^-16

6 0

3 years ago