Answer:
Density of iron is higher than the density of oxygen because iron is solid and the particles in it are very compact but oxygen is not dense as it is gas and particles in it are loosely packed.
Answer:
3MgCl2 has 9 atoms.
Explanation:
The Element Magnesium (Mg) has 3 atoms.
The Element Chloride (Cl) has 6 attoms.
Their fore 6 + 3 is 9 of course. 3MgCl2 has 9 atoms.
BTW: 3MgCl2 is a molecular compound as well as H2O and CO2.
The correct answer is compound
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.
Let:
b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°
We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.
Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.
In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
Answer:
1. Largest force: C; smallest force: B; 2. ratio = 9:1
Explanation:
The formula for the force exerted between two charges is

where K is the Coulomb constant.
q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.
For simplicity, let's combine Kq₁q₂ into a single constant, k.
Then, we can write

1. Net force on each particle
Let's
- Call the distance between adjacent charges d.
- Remember that like charges repel and unlike charges attract.
Define forces exerted to the right as positive and those to the left as negative.
(a) Force on A

(b) Force on B

(C) Force on C

(d) Force on D

(e) Relative net forces
In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

2. Ratio of largest force to smallest
