Answer:
d. Hydrophobic molecules are attracted to each other.
Explanation:
The term “hydrophobic effect” is associated with the spontaneous tendency of macromolecules, such as proteins, to prefer a conformation in an aqueous medium, with hydrophobic groups facing the interior of the mac romolecule, favoring attractive intramolecular interactions, and hydrophilic groups exposed on the surface, for maximize interactions with water molecules in the medium. This is because the hydrophobic molecules are attracted to each other, allowing them to turn inward.
Answer:
1.332 g.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):
<em>∴ (n) of CO₂ = (n) of C₂H₆</em>
<em></em>
∵ n = mass/molar mass
<em>∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆</em>
mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.
mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.
<em>∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆</em> = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) =<em> 1.332 g.</em>
<em></em>
Answer:
2H₂ + O₂ → 2H₂O
Explanation:
Chemical equation:
H₂ + O₂ → H₂O
Balance chemical equation:
2H₂ + O₂ → 2H₂O
Step 1:
H₂ + O₂ → H₂O
Left hand side Right hand side
H = 2 H = 2
O = 2 O = 1
Step 2:
H₂ + O₂ → 2H₂O
Left hand side Right hand side
H = 2 H = 4
O = 2 O = 2
Step 3:
2H₂ + O₂ → 2H₂O
Left hand side Right hand side
H = 4 H = 4
O = 2 O = 2
Answer:
0.0010 mol·L⁻¹s⁻¹
Explanation:
Assume the rate law is
rate = k[A][B]²
If you are comparing two rates,
![\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7Brate%7D_%7B2%7D%7D%7B%5Ctext%7Brate%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7Bk_%7B2%7D%5Ctext%7B%5BA%5D%7D_2%5B%5Ctext%7BB%5D%7D_%7B2%7D%5E%7B2%7D%7D%7Bk_%7B1%7D%5Ctext%7B%5BA%5D%7D_1%5B%5Ctext%7BB%5D%7D_%7B1%7D%5E%7B2%7D%7D%3D%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%5Cright%20%29%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%5Cright%20%29%5E%7B2%7D)
You are cutting each concentration in half, so
![\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctext%7B%20and%20%7D%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%3D%20%5Cdfrac%7B1%7D%7B2%7D)
Then,
