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julsineya [31]
2 years ago
6

PLEASE HELP !!

Physics
1 answer:
Naily [24]2 years ago
4 0

Answer:

Two orbitals for their electrons and six in the 2p subshell

Explanation:

Hope this helps :)

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AysviL [449]

the ansnwer is ATP and ADP whitch is why we need food

8 0
3 years ago
Chemical testing of water
NNADVOKAT [17]
Tap water isn't pure. it just isn't H₂O.
4 0
3 years ago
How far from a -7.80 μC point charge must a 2.40 μC point charge be placed in order for the electric potential energy of the pai
Naddik [55]

Answer:

Distance between two point charges, r = 0.336 meters

Explanation:

Given that,

Charge 1, q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C

Charge 2, q_2=2.4\ \mu C=2.4\times 10^{-6}\ C

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

U=k\dfrac{q_1q_2}{r}

r=k\dfrac{q_1q_2}{U}

r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.

8 0
3 years ago
Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction
lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Cu^{2+}/Cu]}=0.34V

E^o_{[Ag^{+}/Ag]}=0.80V

E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}

E^o=0.80V-(0.34V)=0.46V

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}

E_{cell}=0.434V

Therefore, the cell potential for this cell 0.434 V

8 0
3 years ago
A diode, which allows current to flow in one direction only, consists of two types of semiconductors joined together.
Nata [24]

Answer:

True.

Explanation:

A diode, which allows current to flow in one direction only, consists of two types of semiconductors joined together.

A semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity. Semiconductors are classified into two main categories;

1. Extrinsic semiconductor.

2. Intrinsic semiconductor.

An intrinsic semiconductor is a crystalline solid substance that is in its purest form and having no impurities added to it. Examples of intrinsic semiconductor are Germanium and Silicon.

In an intrinsic semiconductor, the number of free electrons is equal to the number of holes. Also, in an intrinsic semiconductor the number of holes and free electrons is directly proportional to the temperature; as the temperature increases, the number of holes and free electrons increases and vice-versa.

In an intrinsic semiconductor, each free electrons (valence electrons) produces a covalent bond.

6 0
3 years ago
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