The function of a specific enzyme, a certain biological catalyst is most directly influenced by its shape and structure. The shape of the enzyme, primarily determined by the composition of amino acids that constitute or make up the primary structure of it. Will vary in terms of the interactions between these amino acids, and various parts within it. Observed in later levels of protein organization.
Answer:
S₁₂
Explanation:
The freezing point depression (ΔTf) is a colligative property that can be calculated using the following expression.
ΔTf = Kf × m
where,
Kf: freezing point depression
m: molality
ΔTf = Kf × m
m = ΔTf / Kf
m = 0.156 °C / (29.8 °C/m)
m = 5.23 × 10⁻³ m
The molality is:
m = moles of solute / kilograms of solvent
moles of solute = m × kilograms of solvent
moles of solute = 5.23 × 10⁻³ mol/kg × 0.5000 kg
moles of solute = 2.62 × 10⁻³ mol
1.00 g corresponds to 2.62 × 10⁻³ moles. The molar mass of Sₙ is:
1.00 g/2.62 × 10⁻³ mol = 382 g/mol
We can calculate n.
n = molar mass of Sₙ / molar mass of S
n = (382 g/mol) / (32.0 g/mol)
n = 11.9 ≈ 12
The molar formula is S₁₂.
The balanced reaction: C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
We first convert volume of C2H6 to no. of moles. We use the conditions at STP where 1 mol = 22.4 L thus,
Moles C2H6 = 16.4 L/ 22.4 L =0.7321 mol
In order to determine the limiting reagent, we look at the given amounts of the reactants.
0.7321 mol C2H6 (7/2 mol O2 / 1 mol C2H6) = 2.562 mol O2
From the given amounts of the reactants, we can say that O2 is the limiting reactant since we need 2.562 mol O2 to completely react the given amount of C2H6. The excess reagent is C2H6
To calculate for the amount of products and excess reactants:
0.980 mol O2 (2 mol CO2 / (7/2 mol O2)) = 0.56 mol CO2 (22.4 L / 1 mol ) =12.544 L CO2
<span>0.980 mol O2 (1 mol C2H6 / (7/2 mol O2)) = 0.28 mol C2H6
Excess C2H6 = 0.7321 mol - 0.28 mol C2H6 = 0.4521 mol C2H6
We then use the molecular weight of C2H6 to convert the excess amount to grams.
0.4521 mol C2H6 (30.08 g C2H6 / 1 mol C2H6) = <span>13.60 g C2H6
</span></span>
<span>Since the limiting reagent is O2 there will be no oxygen atoms that will be left after the reaction.</span>
Answer:
8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt
Explanation:
Let x represent the liters of the first salt solution
Let y represent the liters of the second salt solution
x + y= 20......equation 1
x= 20-y
The first solution is 20% salt and the second solution is 45% salt
20/100x + 45/100 y= 30/100 × 20
0.2x + 0.45y= 0.3×20
0.2x + 0.45y= 6............equation 2
Substitute 20-y for x in equation 2
0.2(20-y) + 0.45y= 6
4-0.2y+ 0.45y= 6
4 + 0.25y= 6
0.25y= 6-4
0.25y= 2
y= 2/0.25
y= 8
Substitute y for 8 in equation 1
x + y= 20
x + 8=20
x= 20-8
x= 12
Hence 8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt.
