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gtnhenbr [62]
3 years ago
8

The measurement 0.025563 g should be reported as?

Physics
1 answer:
Tomtit [17]3 years ago
7 0
The given mass is 0.025563 g.

Examine the given choices.
a. 0.026 g
This uses 2 significant digits, with rounding to the 3rd decimal place.

b. 2.5 x 10² g = 250 g.
It is incorrect.

c. 0.025 g.
This uses 2 significant digits. It is inaccurate because it does not use rounding to the 3rd decimal place.

d. 0.02 g
This uses one significant digit. It is incorrect for representing the given data.

Answer: a.  0.026 g
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If an atom contains 13 protons, then it has (2.4)a.13 electrons. b. 26 electrons. c. 13 neutrons. d.26 neutrons.
Yuliya22 [10]

If an atom contains 13 protons, then it has <u>13 electrons.</u>

7 0
3 years ago
A department store sells an ""astronomical telescope"" with an objective lens of 30 cm focal length and an eyepiece lens of foca
Yuliya22 [10]

Answer:

The magnifying power of this telescope is (-60).

Explanation:

Given that,

The focal length of the objective lens of an astronomical telescope, f_o=30\ cm

The focal length of the eyepiece lens of an astronomical telescope, f_e=5\ mm=0.5\ cm

To find,

The magnifying power of this telescope.

Solution,

The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :

m=\dfrac{-f_o}{f_e}

m=\dfrac{-30}{0.5}

m = -60

So, the magnifying power of this telescope is 60. Therefore, this is the required solution.

7 0
3 years ago
The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 9.20 nm thick. What is the electric field stren
Sindrei [870]

Answer:

7.60× 10^6 V/m

Explanation:

electric field strength can be determined as ratio of potential drop and distance, I.e

E=V/d

Where E= electric field

V= potential drop= 74.0 mV= 0.07 V

d= distance= 9.20 nm = 9.2×10^-9 m

Substitute the values

E= 0.07/ 9.2×10^-9

= 7.60× 10^6 V/m

5 0
3 years ago
the atomic weight of antimony is 121.76 amu. two naturally occurring isotopes of antimony. 121-Sb has an isotopic mass of 120.90
liubo4ka [24]
Isotope has 57.40% abundance
the other one 100% - 57.40% = 42.60%

mass of 121 (% abundance) + mass of 123 (% abundance) =  total mass (atomic weight)
(120.9038 amu)(0.5740) + (x amu)(0.426) = 121.76 amu
x = 122.9136

hope this help
6 0
3 years ago
¿Cuál es la masa de 12 cm^3 de glicerina, si su densidad absoluta es de 1200kg/m^3? Dé el resultado en gramos
victus00 [196]
La respuesta es 14.4 gramos
7 0
3 years ago
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