Explanation:
As the charge of all electrons are equal, the repulsive force exerted by each of them is also going to be equal. So, as K has more electrons repulsing its valence electron than Na, it has greater electron shielding.
Mass of Oxygen: 0.0159 grams
Moles of Oxygen: 9.94x10^-4
To find the mass of oxygen, subtract the mass of copper from the total mass.

There are 0.0159 grams of Oxygen.
To find how many moles there are, divide the given amount of oxygen by the molar mass (atomic mass) of oxygen because that mass is the same as one mole of oxygen.
Molar mass of Oxygen: 16.00

There are 9.94*10^-4 moles of Oxygen.
Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT. Substituting a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
There are some standard numbers that help us describe the structure of an atom and help us categorize them. Those are the atomic number, the mass number and the numbers of electrons in an atom (or ion). Atoms are electrically neutral, hence they have the same number of protons as electrons. If an atom has a charge and has thus become an ion, it is because electrons joined it or left. For example in this case, since the ion has +2 charge, 2 electrons left it and thus the ion has 4 electrons (2 electrons less than its protons). The mass number is the sum of the protons and neutrons of an atom (that are in the nucleus). In this case, this yields a mass number of 13 for this ion. The atomic number of an atom (or ion) is the total number of protons in the nucleus. Protons do not leave the nucleus except for radioactive reactions and thus the atomic number of an atom (or ion) does not change in chemical reactions. In this case, the ion has an atomic number of 6.
Answer:
2 elements
Explanation:
Octane is composed of carbon and hydrogen atoms.