Answer:
There is 5.56 g of gold for every 1 g of chlorine
Explanation:
The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction 
You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:
The proportion is the equal relationship that exists between two reasons and is represented by: 
This reads a is a b as c is a d.
To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:
Solving for the mass of gold gives:
mass of gold= 5.56 grams
So, <u><em>there is 5.56 g of gold for every 1 g of chlorine</em></u>
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
The correct answer is B. Low chemical reactivity
Answer:
a. the 3 represents the principal energy level
Explanation:
3 is the principal energy level. The p is the sublevel. 4 is the possible occupying electron.
Answer:
Carbon, germanium, tin and lead.
Explanation:
The silicon is belong to the carbon family. There are five elements in carbon family carbon, silicon, germanium, tin and lead. These five elements are present in same group i.e group fourteen. The elements present in same group have same number of valance electrons.
For example.
Carbon electronic configuration:
C₆ = [He] 2s² 2p²
Silicon electronic configuration:
Si₁₄ = [Ne] 3s² 3p²
Germanium electronic configuration:
Ge₃₂ = [Ar] 3d¹⁰ 4s² 4p²
Tin electronic configuration:
Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²
Lead electronic configuration:
Pb₈₂ = [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p²
we can see that in case of all elements there are four valance electrons, which are equal to the valance electrons of silicon.
