Answer:
Are transferred completely from the valence shell of an element to the other
Explanation:
Basically, to form a chemical bond, you either transfer or you share. When you share, it is a case of covalent bonding which can be in several other forms. When there is a transfer, it is a case of ionic bonding.
The basic explanation for this is that while some atoms are electronically sufficient, some are electronically deficient. This means while some atoms are having an excess number of electrons, then some are having less number of electrons.
To satisfy both parties, there must be a transfer if electrons between the two parties. While the one with the excess numbers serves as the donor, the one with insufficient number of electrons serve as the acceptor
Answer:
After the ejection of an alpha particle, the remaining nucleus has a mass number that is four less and an atomic number that is two less, so alpha decay is a type of nuclear fission.
The unit 'mW' means milliwatts. It is a unit of work. There are 1,000 milliwatts in a 1 Watt of work. In 4 hours, there are 14,400 seconds.
Work= Energy/time
17 mW * 1 W/1000 mW = Energy/(14,400 seconds)
Solving for energy,
Energy = 244.8 J
Energy/photon = 244.8 J/(6.04×10²⁰) = 4.053×10⁻¹⁹ J/photon
Using the Planck's equation:
E = hc/λ
where h = 6.626×10⁻³⁴ m²·kg/s, c = 3,00,000,000 m/s and λ is the wavelength
4.053×10⁻¹⁹ J/photon = (6.626×10⁻³⁴ m²·kg/s)(3,00,000,000 m/s)/λ
λ = 4.9×10⁻⁷ m or 49 micrometers
Answer:
Consequently, what happens when gas obtained by heating slaked lime and ammonium chloride is passed through copper sulphate solution? The HCl in the gas mixture will form hydrochloric and the H+ will react with some of the NH3(aq), forming NH4^+, and with some of the SO4^2-, forming HSO4^-.
Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol
