Answer:
There are other details missing in the question. i.e Assume that x is much larger than the separation d between the charges in the dipole, so that the approximate expression for the electric field along the dipole axis E = p/2πε0y3 can be used, where p is the dipole moment, and y is the distance between ions. A) What is magnitude______N B) Direction? +x-direction or -x-direction C) Is this force attractive or repulsive?
A) Magnitude of electric force = 6.576 x 10 raised to power -13 N
B) Since the force direction is always dependent on the electric field and electric field = F/q, since the chlorine has a negative charge as such the direction of the electric force will be in the X - direction
C) Since the charges are of different nature, as such the force between them will be ATTRACTIVE.
Explanation:
The detailed steps is shown in the attachment
Kc= concentration of product divided by concentration of reactant
NO + NO2 ----> N2O3
Kc =(N2O3) / (No)(NO2)
Kc= ( 1.3 )/{ (3.9)(3.8) }
Kc=0.088 ( answer B)
Answer:
The elements become less reactive.
Explanation:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction and reactivity increases because of greater electron affinity.
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased. The electron affinity decreases because of shielding effect and thus atom become less reactive.
Answer:
=> 2.8554 g/mL
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 16.59 g
Volume (v) = 5.81 mL
From our question, we are to determine the density (rho) of the rock.
The formula:

Substitute the values into the formula:

= 2.8554 g/mL
Therefore, the density (rho) of the rock is 2.8554 g/mL.