Answer:
c. reduces the concentration of the hazardous material in the air.
Explanation:
Pollution can be defined as the physical degradation or contamination of the environment through an emission of harmful, poisonous and toxic chemical substances.
Particulate population is a form of pollution that is responsible for the degradation of the environment.
Particulate matter is also referred to as particle pollution or atmospheric aerosol particles and it can be defined as a complex microscopic mixture of liquid droplets and solid particles that are suspended in air. Other forms of particle pollution includes space debris and marine debris.
Some examples of particulate pollution are dusts, soot, dirt, smoke, etc.
Basically, various anthropogenic activities such as construction and agriculture are primary sources of particulate matter because they're capable of causing particle pollution on their own. The other sources of particle pollution is the secondary source which includes factories, cars, trucks, etc.
Vapor dispersion can be defined as a process which is typically used for removing particle pollutants from the atmosphere through the use of vapor or steam.
Hence, vapor dispersion when adopted, reduces the concentration of the hazardous material such as soot, dusts, smoke, etc., in the air.
Potassium (K), Sodium(Na), Iron(Fe), Copper(Cu), Silver(Ag), Tin(Sn), Antimony(Sb), Tungsten(W), Gold(Au), Mercury(Hg), Lead(Pb)
Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
Answer:
1. 58.5g/mol
2. 261g/mol
3. 158g/mol
4. 71g/mol
5. 44g/mol
Explanation:
The molar mass of a compound is the total mass of the sum of masses of all individual elements that make up the compound. First, we need to know the atomic masses of each element in a compound.
1. NaCl
Where; Na = 23, and Cl = 35.5
Molar mass of NaCl = 23 + 35.5
= 58.5g/mol
2. Ba(NO3)2:
Where; Ba = 137, N = 14, O = 16
Molar mass of Ba(NO3)2: 137 + {14 + 16(3)} 2
137 + (14 + 48)2
137 + (62)2
137 + 124
= 261g/mol
3. K(MnO4)
Where; K = 39, Mn = 55, O = 16
Molar mass of KMnO4 = 39 + 55 + 16(4)
= 94 + 64
= 158g/mol
4. Cl2
Where; Cl = 35.5
Molar mass of Cl2 = 35.5(2)
= 71g/mol
5. CO2
Where; C = 12, O = 16
Molar Mass of CO2 = 12 + 16(2)
= 12 + 32
= 44g/mol
