Answer:
1.8 g
Explanation:
Step 1: Write the balanced equation
CH₃CH₃(g) + 3.5 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:112.0 = 0.2684:1.
The experimental mass ratio of CH₃CH₃ to O₂ is 0.60:3.52 = 0.17:1.
Thus, the limiting reactant is CH₃CH₃
Step 3: Calculate the mass of CO₂ produced
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:88.02.
0.60 g CH₃CH₃ × 88.02 g CO₂/30.06 g CH₃CH₃ = 1.8 g
Answer: The molecular formula is 
Explanation:
We are given:
Mass of 
= 0.1605 g
Mass of 
= 0.0220 g
mass of 
= 0.1425 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H =
Moles of S =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = 
For H = 
For S =
The ratio of C : H: S= 3: 5: 1
Hence the empirical formula is 
The empirical weight of = 3(12)+5(1)+1(32)= 73g.
The molecular weight = 146 g/mole
Now we have to calculate the molecular formula.
The molecular formula will be=
Hey There!:
The density is the quotient between the mass of a material and the volume occupied by it, The density can be expressed for a substance or for a mixture of substances. For example, water density in ambient conditions is equal to 1.00 g/cm3, which means that in 1 cm³ or 1 mL, there are 1.0 g of water. Therefore:
D = m / V
D = 20 g / 5 mL
D = 4,0 g/mL
hope that helps!
Answer:
to collect measurable, empirical evidence in an experiment related to a hypothesis the results aiming to support or contradict a theory.
Explanation:
