The best position for the person would be outside, under a clear sky, standing up. He should do it sometime between sunset and sunrise, from a day before until a day after the moment of Full Moon.
Answer:
The leaves of the electroscope move further apart.
Explanation:
This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.
Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.
So, the leaves move further apart.
Answer:
1.90×10²⁰ Electrons
Explanation:
From the question,
Q = It.................... Equation 1
Where Q = charge flowing through the wire, I = current, t = time
Given: I = 4.35 A, t = 7.00 s
Substitute these values into equation 1
Q = 4.35(7.00)
Q = 30.45 C.
But,
1 electron contains 1.6×10⁻¹⁹ C
therefore,
30.45 C = 30.45/1.6×10⁻¹⁹ electrons
= 1.90×10²⁰ Electrons
Complete question:
Consider the hypothetical reaction 4A + 2B → C + 3D
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?
Answer:
the final concentration of A is 0.992 M.
Explanation:
Given;
time of reaction, t = 4.0 s
rate of change of the concentration of B = -0.0760 M/s
initial concentration of A = 1.600 M
⇒Determine the rate of change of the concentration of A.
From the given reaction: 4A + 2B → C + 3D
2 moles of B ---------------> 4 moles of A
-0.0760 M/s of B -----------> x
⇒Determine the change in concentration of A after 4s;
ΔA = -0.152 M/s x 4s
ΔA = -0.608 M
⇒ Determine the final concentration of A after 4s
A = A₀ + ΔA
A = 1.6 M + (-0.608 M)
A = 1.6 M - 0.608 M
A = 0.992 M
Therefore, the final concentration of A is 0.992 M.
Answer:
B) protons and neutrons.
Explanation:
The protons and neutrons are located in the nucleus of the atom and represent most of the 'mass' of the atom, that's their count that determine the 'mass' of an atom (like 12 for Carbon).
The electrons rotate around the nucleus and have a negligible mass.
