Answer:
B. -2
Explanation:
The total charge on an atom is the sum of all individual charges present in it. Therefore, the total charge on this atom is given by the following formula:

where,
q = total charge on atom = ?
= no. of protons in the atom = 15
= no. of electrons in the atom = 17
= no. of neutrons in the atom = 12
= charge on proton = +1
= charge on electron = -1
= charge on neutron = 0
Therefore,

Hence the correct option is:
<u>B. -2</u>
Due to the theory of dried enzimes and philosophy, the spilled milk would have an higher entropy.
Answer:

Explanation:
Hello.
In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

The distance is clearly 1743 km and the time is:

Thus, the velocity turns out:

Which is a typical velocity for a plane to allow it be stable when flying.
Best regards.
Answer:
6.86 × 10²⁴ kg
Explanation:
The mass of the earth m = density of earth, ρ × volume of earth, V
m = ρV
The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m
Since m = ρV
m = ρ4πr³/3
So, substituting the values of the variables into the equation for the mass of the earth, m, we have
m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3
m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3
m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3
m = 6546105.64378π × 10¹⁸ kg/3
m = 20565197.400122 × 10¹⁸ kg/3
m = 6855065.8 × 10¹⁸ kg
m = 6.8550658 × 10²⁴ kg
m ≅ 6.86 × 10²⁴ kg