Answer:
6.564×10¹⁶ fg.
Explanation:
The following data were obtained from the question:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt in fg =?
Next, we shall determine the mass of the salt in grams (g). This can be obtained as follow:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt =?
Mass of salt = (Mass of beaker + salt) – (Mass of beaker)
Mass of salt = 142.54 – 76.9
Mass of salt = 65.64 g
Finally, we shall convert 65.64 g to femtograms (fg) as illustrated below:
Recall:
1 g = 1×10¹⁵ fg
Therefore,
65.64 g = 65.64 g × 1×10¹⁵ fg / 1g
65.64 g = 6.564×10¹⁶ fg
Therefore, the mass of the salt is 6.564×10¹⁶ fg.
Answer:
The substance is a mixture because it left a white powder in the beaker when boiled.
Explanation:
Answer:
(i) Bohr; (ii) de Broglie; (iii) Heisenberg (v) Schrödinger
Explanation:
(i) Niels Bohr — 1913 — proposed that electrons travel in fixed orbits with <em>quantized energy levels</em> and that they jump from one energy level to another by absorbing or emitting quanta of light.
(ii) <em>Louis de Broglie</em> — 1924 — proposed the wave nature of electrons and suggested that all matter behaves as both waves and particles (<em>wave-particle duality</em>).
(iii) Werner Heisenberg — 1927 — formulated quantum mechanics in terms of matrices and proposed his famous <em>uncertainty principle</em>.
(v) Erwin Schrödinger — 1926 — applied wave mechanics to the electron in a hydrogen atom, showing that electrons exist in <em>orbitals </em>rather that orbits.
(iv) <em>Ernest Rutherford</em> — 1911 — proposed that atoms have most of their mass in a central nucleus (<em>nuclear atom</em>). Quantum mechanics had not yet been invented.
Answer:
B. Charges ( a slight positive charge on one end, and a slight negative charge on the other).
Answer:

Explanation:
Hello there!
In this case, according to the given chemical reaction for this problem about stoichiometry:

Whereas there is a 3:2 mole ratio of oxygen (molar mass = 32.0 g/mol) to iron (III) oxide (molar mass = 159.69 g/mol) and therefore, the correct stoichiometric setup is:

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