Answer:
The volume of gold is 3.3 cm³.
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m³.
Other units are given below,
g/cm³, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Given data:
Mass of gold = 63 g
Density of gold = 19.32 g/cm³
Volume of gold = ?
Solution:
d = m/v
v = m/d
v = 63 g/ 19.32 g/cm³
v = 3.3 cm³.
Explanation:
Let us assume that the solution contains both and which has a cell voltage of 0.719 V.
Therefore, voltage of cell contains both Ag/AgCl reference electrode where electrode is 0.719 V.
As, = 0.719 V
It is known that potential of the silver-silver chloride reference electrode is 0.197 V.
Hence, = 0.197 V. Now, calculate as follows.
= 0.719 V
= 0.719 V
= 0.916 V
Now, voltage of the cell that contains both calomel reference electrode and electrode as follows.
= calomel electrode = 0.241 V
Voltage of cell =
= 0.916 V - 0.241 V
= 0.675 V
Thus, we can conclude that 0.675 V is the new voltage.
It may seem remarkable that we can learn about the composition of distant stars by studying the light they emit. In fact, we can learn a great deal, not only about the chemical elements present, but also about physical conditions. The key is to spread the light out by color, producing a spectrum like the one shown in Fig. 1. This lab explores some of the basic ideas used to analyze spectra.
First question. Applying ideal gas equation PV=nRT, P= 101.3 x 10³Pa = 1atm. therefore, 1 x 260 x 10^-3 = n x 0.082 x 294.( Temperature in kelvin=273+21). n = 0.01 moles. Volume of gas at STP= n x 22.4 = 0.01x22.4 = 0.224L. Hope this helps