No, because a child under the age of 16 does not have the right to engage in such relationships and it's punishable by law if the relationship becomes intimate.
<span>Tides do place a drag on the earth's rotation, but gradually. This energy lost from the earth's rotation is picked up by the moon's orbit around the earth. The moon is receding from the earth. So, in answering your question, the length of the day has been increasing. Without getting into the arduous math involved, in the very distant future the moon will have reached its max distance and the earth's rotation will be in step with the moon's orbit.</span>
Answer:
The mass of copper sulfate in 25 cm³ of the copper sulfate solution is 0.45 g
Explanation:
The given parameters are;
The volume of the copper sulfate solution = 100 cm³
The mass of the copper sulfate in the solution = 1.8 g
Therefore, the mass of copper sulfate in 25 cm³ of the solution is given as follows;
The mass of copper sulfate in 100 cm³ of the solution = 1.8 g
The mass of copper sulfate in 1 cm³ of the solution = 1.8 g/100 = 0.018 g
Therefore;
The mass of copper sulfate in 25 × 1 = 25 cm³ of the solution, m = 25×0.018 g = 0.45 g
∴ The mass of copper sulfate in 25 cm³ of the solution, m = 0.45 g
Answer:
0.38 moles He
Explanation:
from the ideal gas law PV= nRT you can calculate the volume of 1 mole at STP (273..1K and 1 atm of pressure) the MOLAR VOLUME IS
22.4 L
In the problem you have 8.4L
8.4 L/22.4 L=0.375 moles
so
0.38 moles He
the force between the electron and the proton.
a) Use F = k * q1 * q2 / d²
where k = 8.99e9 N·m²/C²
and q1 = -1.602e-19 C (electron)
and q2 = 1.602e-19 C (proton)
and d = distance between point charges = 0.53e-10 m
The negative result indicates "attraction".
the radial acceleration of the electron.
b) Here, just use F = ma
where F was found above, and
m = mass of electron = 9.11e-31kg, if memory serves
a = radial acceleration
the speed of the electron.
c) Now use a = v² / r
where a was found above
and r was given
<span> the period of the circular motion.</span>
d) period T = 2π / ω = 2πr / v
where v was found above
and r was given