(a) Interpretation: The number of moles of platinum in the cone should be calculated. Concept introduction: The number of moles of a substance is related to molar mass of the substance as follows: n = m M Here, m is mass and M is molar mass. In 1 mol of a substance there are 6.023 × 10 23 units of that substance. Also, density of a substance is related to mass and volume as follows: d = m V Here, m is mass and V is volume of the substance.
(a) Interpretation: The number of moles of platinum in the cone should be calculated. Concept introduction: The number of moles of a substance is related to molar mass of the substance as follows: n = m M Here, m is mass and M is molar mass. In 1 mol of a substance there are 6.023 × 10 23 units of that substance. Also, density of a substance is related to mass and volume as follows: d = m V Here, m is mass and V is volume of the substance.
Definition Definition Number of atoms/molecules present in one mole of any substance. Avogadro's number is a constant. Its value is 6.02214076 × 10 23 per mole.
Chapter 3, Problem 6QAP
Interpretation Introduction
(a)
Interpretation:
The number of moles of platinum in the cone should be calculated.
Concept introduction:
The number of moles of a substance is related to molar mass of the substance as follows:
n=mM
Here, m is mass and M is molar mass.
In 1 mol of a substance there are 6.023×1023 units of that substance.
Also, density of a substance is related to mass and volume as follows:
d=mV
Here, m is mass and V is volume of the substance.
Interpretation Introduction
(b)
Interpretation:
The number of electrons in there in the cone should be calculated.
Concept introduction:
The number of moles of a substance is related to molar mass of the substance as follows:
n=mM
Here, m is mass and M is molar mass.
According to Avogadro’s law, in 1 mol of a substance there are 6.023×1023 units of that substance.
With the advent of techniques such as scanning tunneling microscopy, it is now possible to “write” with individual atoms by manipulating and arranging atoms on an atomic surface.(A) If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? Show all work. [2](B) If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part (B) is used, what number of ruthenium atoms is needed to construct the surface? Show all work. [2]
How many atoms of O are found in 1.00 x 10 2 g of K 2Cr 2O 7 ?
The density of silver is 10.5 g/cm^3. How many silver atoms are present in a silver bar that measures 0.100 m x 0.050 m x 0.010 m?