Answer:
Blue dye spreads out quicker
Explanation:
This is because the dye gets spread out by water molecules. Warm water has faster moving water molecules than cold water.
Rutherford theorized that atoms have their charge concentrated in a very small nucleus.
This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).
Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.
That is different from Plum Pudding model, because it shows that most of the atom is empty space.
According to Rutherford model of the atom:
1) Atoms have their charge concentrated in a very small nucleus.
2) Major space in an atom is empty.
3) Atoms nucleus is surrounded by negatively charged particles called electrons.
4) An atom is electrically neutral.
The mass of melted gold to release the energy would be 3, 688. 8 Kg
<h3>How to determine the mass</h3>
The formula for quantity of energy is given thus;
Q = n × HF
Where n represents number of moles
HF represents heat of fusion
To find the number of moles, we have
235.0 = n × 12.550
number of moles =
= 18. 725 moles
Note that molar mass of Gold is 197g/ mol
Let's note that;
Number of moles = mass/ molar mass
Mass = number of moles × molar mass
Mass = 18. 725 × 197
Mass = 3, 688. 8 Kg
Thus, the mass of melted gold to release the energy would be 3, 688. 8 Kg
Learn more about molar heat of fusion here:
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Answer:
The ground state configuration is the lowest energy, most stable arrangement. An excited state configuration is a higher energy arrangement (it requires energy input to create an excited state). Valence electrons are the electrons utilised for bonding.
or the
FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.
Element Atomic number Electron configuration
sulfur 16 1s22s22p63s23p4
chlorine 17 1s22s22p63s23p5
argon 18 1s22s22p63s23p6
or the
Two electrons
Two electrons fill the 1s orbital, and the third electron then fills the 2s orbital. Its electron configuration is 1s22s1.
Explanation:
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Answer:
The concentration of the solution will be much lower than 6M
Explanation:
To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.
From
n= CV
n = number of moles m/M( m= mass of solid, M= molar mass of compound)
C= concentration of substance
V= volume of solution
m=120g
M= 40gmol-1
V=500ml
120/40= C×500/1000
C= 120/40× 1000/500
C=6M
This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.
This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.