How many ml of naoh are needed to neutralize 15.0 ml of 0.350m ch3cooh?

Elis [28]

Explanation:

The ionization energy of an atom is the amount of energy that is required to remove an electron from a mole of atoms in the gas phase:

M(g) ® M+(g) + e-

It is possible to remove more electrons from most elements, so this quantity is more precisely known as the first ionization energy, the energy to go from neutral atoms to cations with a 1+ charge. The second ionization energy is the energy that is required to remove a second electron, to form 2+ cations from 1+ cations:

M+(g) ® M2+(g) + e-

The third ionization energy is the energy required to form 3+ cations:

M2+(g) ® M3+(g) + e-

and so on. Ionization energies are always positive numbers, because energy must be supplied (an endothermic energy change) to separate electrons from atoms. The second ionization energy is always larger than the first ionization energy, because it requires even more energy to remove an electron from a cation than it is from a neutral atom.

The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy, while francium has one of the lowest.

From top to bottom in a group, orbitals corresponding to higher values of the principal quantum number (n) are being added, which are on average further away from the nucleus. Since the outermost electrons are further away, they are less strongly attracted by the nucleus, and are easier to remove, corresponding to a lower value for the first ionization energy.From left to right across a period, more protons are being added to the nucleus, but the number of electrons in the inner, lower-energy shells remains the same. The valence electrons feel a higher effective nuclear charge — the sum of the charges on the protons in the nucleus and the charges on the inner, core electrons. The valence electrons are therefore held more tightly, the atom decreases in size (see atomic radius), and it becomes increasingly difficult to remove them, corresponding to a higher value for the first ionization energy.

The following charts illustrate the general trends in the first ionization energy:

Dunno kung tama beng pero trysorry kung mali

8 0

3 years ago

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Nana76 [90]

Answer:

Therefore, when an atom becomes a positive ion is pulls its electrons closer, decreasing is radius moreover, when it becomes a negative ion, it pulls its electrons closer and decreases the radius.

5 0

2 years ago

Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of CO2 and 11.5 g of H2O. What is the empirical formula for the h

cluponka [151]

<span>C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5</span>

3 0

3 years ago

What are 3 household substances that contain acids and say why are they useful

Nonamiya [84]

Answer:

battery. you use it for your remotes. vinegar. you use vinegar for cooking of cleaning. nail polish remover. you use this for removing nail polish

6 0

2 years ago

How many moles are in 558kg of dinitrogen dioxide?

ella [17]

How many grams N2O in 1 mol? The answer is 44.0128. We assume you are converting between grams N2O and mole. You can view more details on each measurement unit: molecular weight of N2O or mol This compound is also known as Nitrous Oxide.

4 0

3 years ago