Answer is: <span>the mass of the excess reactant (ethane) leftover is 90.135 grams.
</span>Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O<span>(g).
m(</span>C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.
Answer:
6.05 g
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity
V = volume of solution in liter ,
n = moles of solute ,
From the question ,
M = 200mM
Since,
1 mM = 10⁻³ M
M = 200 * 10⁻³ M
V = 250 mL
Since,
1 mL = 10⁻³ L
V = 250 * 10⁻³ L
The moles can be calculated , by using the above relation,
M = n / V
Putting the respective values ,
200 * 10⁻³ M = n / 250 * 10⁻³ L
n = 0.05 mol
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
m = 121 g/mol
n = 0.05 mol ( calculated above )
The mass of tri base can be calculated by using the above equation ,
n = w / m
Putting the respective values ,
0.05 mol = w / 121 g/mol
w = 0.05 mol * 121 g/mol
w = 6.05 g
The environment affects natural selection. Speices need to adapt to changes in environmnet. The weaker organisms which fail to adapt to environment perish while the stronger ones who adapt to the environment survive. This is how environment affects natural selection.
It equals 0.005 but plz mark this as the brainliest question plz
