The answer is going to be mercury
The first thing we need to do here is to recognize the unit of molarity and the units of the given percentage of nitric acid.
Molarity is mol HNO3 / L of solution. This is our aim
The given percentage is 0.68 g HNO3/ g solution
multiplying this with density to convert g solution into mL solution and dividing with the molecular weight of HNO3 (63 g/mol) to convert g HNO3 to mol. Therefore we obtain
0.016 mol/ mL or 16.23 mol/ L (M)
Answer:
Element 1
Explanation:
The ionization energy is defined as the energy required to remove electrons from the atoms.
We know that the nucleus of the atom attracts the electrons, thus, bound these electrons to the atom.
This means that as the radius decreases, the force of attraction between the nucleus and the electron will increase, therefore, the energy required to remove the electron would increase (and vice-versa).
Based on the above, the atom with the smallest radius would be the atom with the largest first ionization energy.
Hope this help :)
Answer:
0.6378 M
Explanation:
Molarity is defined by Moles per liter.
Plugging the given information in, we get (14.968 moles)/(23.47 L) which comes out to be about 0.6378 M
0.300 M IKI represents the
concentration which is in molarity of a potassium iodide solution. This means
that for every liter of solution there are 0.300 moles of potassium iodide. Knowing
that molarity is a ratio of solute to solution.
By using a conversion factor:
100 ml x (1L / 1000 mL) x (0.300
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g
Therefore, in the first
conversion by simply converting the unit of volume to liter, Molarity is in L
where the volume is in liters. The next step is converted in moles from volume
by using molarity as a conversion factor which is similar to how density can be
used to convert between volume and mass. After converting to moles it is simply
used as molar mass of Kl which is obtained from periodic table to convert from
mole to grams.
In order to get the grams of IKI
to create a 100 mL solution of 0.600 M IKI, use the same formula as above:
100 ml x (1L / 1000 mL) x (0.600
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g