Answer:
0.108mol of Hydrogen
Explanation:
The formula for the compound is: CH3COOH
From the formula of the compound,
There are 2moles of oxygen and 4moles of Hydrogen.
If for every 2moles of oxygen, 4moles of Hydrogen is present.
Then, for 0.054 moles of oxygen = (0.054 x 4)/2 = 0.108mol of Hydrogen is present
Answer:
Ek = (RT/zF)*ln ( [k+]o/[K+]i )
Explanation:
R = gas constant (8.31 J/Kmol)
T = Temperature (k)
F = Faraday constant (9.65 * 10exp4 coulomb/mole)
z = valence of the ion (1)
[k+]o = Extracellular K concentration in mM
[K+]i = Intracellular K concentration in mM
ln = logarithm with base e
First, you need to find the number of moles of OH⁻ in a 250mL solution of 0.100M OH⁻. To do this, multiply 0.250L by 0.100M to get 0.025mol OH⁻. Then you use the fact that 1 mole of Sr(OH₂)·8H₂O contains 2 moles of OH⁻ which means that 0.0125mol of Sr(OH)₂·8H₂O contains 0.025mol OH⁻ (0.025/2=0.0125). Then to find the amount of Sr(OH)₂·8H₂O is needed you multiply its molar mass (265.76g/mol) by 0.0125mol to get 3.322g.
Therefore you need 3.322g of Sr(OH)₂·8H₂O.
I hope that helps. Let me know if anything is unclear.
Actually, that does not happen until the protostar becomes a star when nuclear ignition starts and is maintained. It takes awhile for new star to go through its T-Tauri stage and settle down on the main sequence.
<span>A STAR does not reach hydrostatic equilibrium until it on the main sequence. Otherwise, it would remain a brown dwarf with not enough mass to to maintain nuclear fusion for more than 3,000 to 10,00 years. </span>
Explanation:
Metals are the species which readily lose electrons in order to attain stability. This electron lost by the atom is actually present in its outermost shell which is also known as valence shell.
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period.
But when we move down a group then there occurs an increase in atomic size of the atoms due to addition of number of electrons in the atoms. Hence, ionization energy decreases along a group.
Thus, we can conclude that metals have low ionization energies and readily share their valence or outer electrons with each other to form an electron sea. These electrons are delocalized or shared among all the atoms that are bonded together and can therefore move freely throughout the metal structure.