Answer:
Charge the balloon, hold it near an electroscope, and determine if the electroscope leaves move.
Explanation:
The gold leaf electroscope is an instrument used to detect if a body is charged. It has two gold leafs suspended from a brass stem in a vacuumed glass jar and connected to a metal cap(Toppr).
When the test body is allowed to touch the metal cap, a change in the size of the leaves shows whether the body is charged or not.
Since we are suspecting the balloon to be made up of a metal; metals can be charged. We can test if there is really a charge on the balloon by bringing it near an electroscope to see if the electroscope moves.
A. High intermolecular forces of attraction. If there are high intermolecular forces, the molecules will need large energies to escape into the liquid. The substance will nave a high melting point.
The other options are <em>incorrect </em>because they are <em>weak force</em>s. They would cause <em>low melting points</em>.
Answer:
6 x 10⁶ g Fe
Explanation:
Step 1: Set up dimensional analysis
7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)
Step 2: Multiply, divide, and cancel out units
atoms Fe and atoms Fe cancel out.
mol Fe and mol Fe cancel out.
We should be left with g Fe.
7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe
116279(55.85) = 6.49 x 10⁶ g Fe
Step 3: Sig figs
There is only 1 sig fig in this problem.
6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe
Answer:
induced dipole-dipole forces or London Dispersion forces / van der Waals forces.
Explanation:
Hexane is non-polar in nature. This is due to :
The bond in the molecule is C-H, which is non-polar in nature because the carbon and the hydrogen having very similar electronegativity values.
Hexane is also symmetric.
The intermolecular force acting in the molecule of the hexane are induced the dipole-dipole forces or London Dispersion forces / van der Waals forces.
Answer:
C. 1.35
Explanation:
2NH3 (g) <--> N2 (g) + 3H2 (g)
Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0
change in concentration 2x x 3x
-0.84 M +0.42M +1.26M
Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M
concentration
Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M
Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M
Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M
K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M
