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sveta [45]
4 years ago
10

El punto de fusión del plomo es de 327 °C y su punto de ebullición, de 1750 °C. a) ¿En qué estado se encontrará un trozo de plom

o calentado hasta 325 °C?
Physics
1 answer:
Temka [501]4 years ago
6 0

Answer:

01001101 01100101 00100000 01101100 01100001 00100000 01110011 01110101 01100100 01100001 00100000 01110100 01110010 01100101 01110011 00100000 01100011 01101111 01101010 01101111 01101110 01100101 01110011

Explanation: traducelo a binario :)

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What substance can be used to electrolyze water?
lubasha [3.4K]
C/any electrolyte that is not easily reduced or oxidized
6 0
3 years ago
Which of the following is not part of the process known as oxidative phosphorylation?(a) Molecular oxygen serves as a final elec
Andre45 [30]

Answer:

(d) ATP molecules are produced in the cytosol as glucose is converted into pyruvate.

5 0
4 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
4 years ago
How much force is needed to accelerate a 1.5 kg physics book to an acceleration of 6<br> m/s^2?
aleksley [76]

Answer:

Force = 8.0 k g m / s

Explanation:

Force = mass x acceleration

Mass = 4.0 k g Acceleration = 2.0 m / s 2

Hence,force = ( 4.0 x 2.0 ) k g m / s 2 = 8.0 k g m / s 2

3 0
3 years ago
A satellite with mass 6000 kg is orbiting the planet at 2500 km above the planet's
9966 [12]

By the law of universal gravitation, the gravitational force <em>F</em> between the satellite (mass <em>m</em>) and planet (mass <em>M</em>) is

<em>F</em> = <em>G</em> <em>M</em> <em>m</em> / <em>R </em>²

where

<em>• G</em> = 6.67 × 10⁻¹¹ m³/(kg•s²) is the universal gravitation constant

• <em>R</em> = 2500 km + 5000 km = 7500 km is the distance between the satellite and the center of the planet

Solve for <em>M</em> :

<em>M</em> = <em>F R</em> ² / (<em>G</em> <em>m</em>)

<em>M</em> = ((3 × 10⁴ N) (75 × 10⁵ m)²) / (<em>G</em> (6 × 10³ kg))

<em>M</em> ≈ 2.8 × 10¹⁴ kg

6 0
3 years ago
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