Answer:
The most active nonmetals belong to the halogen family, which sits to the left of the noble gases on the right side of the periodic table. The halogens are so reactive that they are never found in nature by themselves. The elements fluorine, chlorine, bromine, iodine and astatine make up the halogen family.
Answer:
4777.09grams
Explanation:
To find the mass of oxygen, we first need to find the mass of potassium sulfate (K2SO4).
Since there are 4.50 x 10^25 formula units of potassium sulfate, we can find the number of moles in K2SO4 by dividing by Avagadros number (6.02 × 10^23 units). That is;
number of moles of K2SO4 (n) = 4.50 x 10^25 ÷ 6.02 × 10^23
= 0.747 × 10^ (25-23)
= 0.747 × 10^2
= 7.47 × 10^1 moles
Mass in grams of K2SO4 can be calculated thus: molar mass of K2SO4 × moles
= 174.252 g/mol × 7.47 × 10^1 moles
= 13016.62grams.
Since mass of oxygen in 1 mol of K2SO4 = O4 = 63.996 g/mol
We find the percentage by mass of oxygen in K2SO4 as follows:
= 63.996/174.252 × 100
= 0.367 × 100
= 36.7% by mass of oxygen.
This means that in 1gram of K2SO4, there are 0.367gram of Oxygen. Hence, in 13016.62grams (4.50 x 10^25 units) of K2SO4, there will be;
0.367 × 13016.62
= 4777.09grams of oxygen.
Answer:
NaCN- basic salt
KNO3 - neutral salt
NH4Cl - acid salt
NaHCO3 - acid salt
Na3PO4 - acid salt
Explanation:
Salt hydrolysis a process by which salts react with water giving an acid and a base.
When we dissolve NaCN in water, we have;
NaCN + - ⇄ Na^+ + CN^-
KNO3 ------> K^+ + NO3^-
NH4Cl ------> NH4^+ + Cl^-
NaHCO3 -----> Na^+ + HCO3^-
Na3PO4 ----> 3Na^+ + PO4^3-
Note that if a salt is formed from a weak acid and a strong base, the salt will be a basic salt e.g NaCN formed from weak HCN and strong NaOH.
If a salt is formed from a strong acid and weak base, the salt will be acidic, e.gNH4Cl formed from weak NH3 and strong HCl.
If a salt is formed from a strong acid and strong base, the salt will be neutral, e.g KNO3 formed from strong KOH and strong HNO3.
We must first write out the entire equation for this reaction which is as follows:
CO + Fe2O3 --> Fe + CO2
Now we must balance this equation which provides us with the following equation:
3 CO + Fe2O3 --> 2Fe + 3 CO2
We are told that we have excess Fe2O3, so that suggests that CO is the limiting reagent. We now simply convert the mass of Fe given to moles of Fe, and convert moles of Fe to moles of CO.
35.0 g Fe/ 55.845 g/mol = 0.627 moles Fe
0.627 moles Fe x (3 moles CO)/(2 moles Fe) = 0.940 moles CO
Now with the moles of CO present, we simply convert this back to mass using the molecular weight of CO.
0.940 moles CO x 28.01 g/mol = 26.3 g CO.
Therefore, 26.3 g of CO are needed to produce 35.0 g of Fe. Since we began with three significant figures in our starting mass, our answer must also have three significant figures.