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ioda
3 years ago
12

The overall reaction in the lead storage battery is: Pb (s) + PbO2 (s) + 2 H+ (aq) + 2 HSO4 - (aq)  2 PbSO4 (s) + 2 H2O (l) Cal

culate the potential at 310 K for this battery when [H+ ] = [HSO4 - ] = 4.5 M. At 310 K E o = 2.08 V for the lead storage battery. (a) 2.10 V (b) 2.24 V (c) 2.16 V (d) 2.12 V (e) not enough inf
Physics
1 answer:
VMariaS [17]3 years ago
4 0

Answer:

d) 2.12 V

Explanation:

E =E° - RT/nF log 1/H⁺ x HSO₄⁻

E = 2.08 - ( .059/2 X log 1/4.5x4.5 )

2.08 - (-0.038 ) = 2.118 =2.12 V

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Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

The time period (T) = \frac{1}{f}

or

T =  \frac{1}{1.03\times 10^8}  = 9.6 x 10⁻⁹ seconds  

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s  

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s  

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}  

Thus, the minimum distance to target = \frac{290}{2} = 145 m

6 0
3 years ago
PLEASE HELP! thxssss<br>​
Cerrena [4.2K]

Answer:

D is correct

Explanation:

5 0
3 years ago
the resistivity of a given wire of cross session area 0.7mm² is 49×10⁴.calcuate the resistance of a 2m lenght of wire​
sp2606 [1]

R = 1.4GΩ.

The relation between the resistance and the resistivity is given by the equation R = ρL/A, where ρ is the resistivity of a given material, L is the length and A is the cross-sectional area of the material.

To calculate the resistance of a wire of L = 2m, ρ = 49x10⁴Ω.m and A = 0.7mm² = 0.7x10⁻³m² we have to use the equation  R = ρL/A.

R = [(49x10⁴Ω.m)(2m)/0.7x10⁻³m²

R = 98x10⁴Ω.m²/0.7x10⁻³m²

R = 1.4x10⁹Ω = 1.4GΩ

6 0
3 years ago
How is it possible for one electrically neutral atom to exert an electrostatic force on other electrically neutral atom?
Free_Kalibri [48]

Answer:

Explained

Explanation:

Although Atom are electrically neutral. But atom atom is combination of nucleus and electrons. The nucleus of the atom is composed neutron and positively charged protons. On the outside of nucleus at some distance are the electrons which are negatively charged. So, there is difference in position of the two differently charge species. So, this way a electrically neutral atom can exert a electrostatic force on other electrically neutral atom

8 0
4 years ago
Two samples of aluminum foil, 27 grams and 54 grams, are cut out from a roll. Which of the following properties is different for
Delicious77 [7]
Before answering this question, you must know the concept between extensive and intensive property. The extensive property does not depend on the amount of substance (like mass), which is the opposite of intensive properties. From the given choices, the rest are extensive properties except for <em>amount of matter</em>. Hence, that is the answer.
4 0
4 years ago
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