Answer;
D. C6H6, C2H5OH, C6H5CH3, and C3H5(NO3)3.
Explanation;
-Organic compound are chemical compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen, oxygen, or nitrogen, like the examples above in choice D. . Inorganic Compoundson the other hand, are compounds made from any elements except those compounds of carbon, such as CaSO4, CaCl2 and Ca3(PO4)2 in A, B and C respectively.
Answer:
The area is 2.5 m²
Explanation:
Step 1: Data given
A car has a 25 m² hydraulic lift platform that weighs 15000 N
The smaller piston required a force to lift = 1/100 its weight
Step 2: Calculate the area of the smaller piston
Pressure = Force / Area
F1 / A1 = F2 / A2
⇒ with F1 = weight of the hydraulic lift = 15000 N
⇒ with A1 = area of the hydraulic lift = 25 m²
⇒ with F2 = 1/100 of it's weight = 1500 N
A2 = F2 / (F1/A1)
A2 = 1500 ( 1500/25)
A2 = 2.5
The area is 2.5 m²
Answer:
Answer is B
Explanation: Trust
Answer:
By measuring the total mass of the matter before as well as after any physical or chemical change.
Explanation:
As mentioned, according to the law of conservation of matter, no atoms can be created or destroyed when under physical or chemical change.
Naturally, all the matters have some mass associated with it.
To verify the law of conservation of matter, at first, measure the mass of matter before it underwent any physical or chemical change. Then, again, measure the mass of the matter after the physical or chemical change.
Now, observe both the mass, both must be equal. This verifies the law of conservation of matter.
Answer:
The % yield of the reaction is 73.8 %
Explanation:
To solve this, we list out the given variables thus
Mass of aluminium in the experiment = 2.5 g
mass of oxygen gas in the experiment = 2.5 g
Molar mass of aluminium = 26.98 g/mol
molar mass of oxygen O₂ = 32 g/mol
The reaction between aluminium and gaseous oxygen can be written as follows
4Al + 3O₂ → 2Al₂O₃
Thus four moles of aluminium forms two moles of aluminium oxide
Thus (2.5 g)÷(26.98 g/mol) = 0.093 mole of aluminium and
(2.5 g)÷(32 g/mol) = 0.078125 moles of oxygen
However four moles of aluminium react with three moles of oxygen gas O₂
1 mole of aluminum will react with 3/4 moles of oxygen O₂ and 0.093 mole of aluminium will react with 0.093*3/4 moles of O₂ = 0.0695 moles of Oxygen hence aluminium is the limiting reagent and we have
1 mole of oxygen will react with 4/3 mole of aluminium
∴ 0.078125 mole of oxygen will react with 0.104 moles of aluminium
Therefore 0.093 mole of aluminium will react with O₂ to produce 2/4×0.093 or 0.0465 moles of 2Al₂O₃
The molar mass of 2Al₂O₃ = 101.96 g/mol
Hence the mass of 0.0465 moles = number of moles × (molar mass)
= 0.0465 moles × 101.96 g/mol = 4.74 g
The of aluminium oxide Al₂O₃ is 4.74 g, but the actual yield = 3.5 g
Therefore the Percentage yield = ×100 = × 100 = 73.8 % yield