Kepler's
third law shows the relationship between the orbital period of an object and
the distance between the object and the object it orbits.
The
simplified version of this law is: P^2 = a^3
Where,
P =
period of the orbit in years = 0.62 years
a =
average distance from the object to the object it orbits in AU. The
astronomical unit AU is a unit of length which is roughly equivalent to the
distance from Earth to the Sun.
Therefore
calculating for a:
0.62
^ 2 = a ^ 3
a =
0.62 ^ (2/3)
a =
0.727 AU = 0.72 AU
Therefore we can interpret this as: The distance from Venus to the Sun is about 72% of the distance from Earth to
Sun.
<span>Answer:
B. 0.72 AU</span>
Answer:
The correct statement is that the point of initial resistance is the level of depression that will fill the pipette with the desired volume of solution.
Explanation:
The pipette can be otherwise stored vertically or horizontally, this has nothing to do with the operation, and before the operation, the dial needs to be set. Also, the plastic pipette tip should not be ejected at all, thus, no new disposable plastic tip is required for each sample.
The correct statement is that the point of initial resistance is the level of depression that will fill the pipette with the desired volume of solution.
3) CH₃-COOH + NH₃ → CH₃-COO⁻NH₄⁺
4) 2 FeCl₃ + 3 Ag₂SO₃ → Fe₂(SO₃)₃ + 6 AgCl
5) 2 Al + 3 NiCl₂ → 2 AlCl₃ + 3 Ni
6) 4 LiCl + Pb(NO₂)₄ → 4 LiNO₂ + PbCl₄
7) 3 H₂SO₄ + 2 Al(OH)₃ → Al₂(SO₄)₃ + 6 H₂O
8) Cd(NO₃)₂ + Na₂S → CdS + 2 NaNO₃
9) Cr₂(SO₄)₃ + 3 (NH₄)₂CO₃ → Cr₂(CO₃)₃ + 3 (NH₄)₂SO₄
Answer:
0.22 mol HClO, 0.11mol HBr.
0.25mol NH₄Cl, 0.12 mol HCl
Explanation:
A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.
Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.
Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em> will produce NH₄⁺. 0.25mol HCl consume all NH₃.
Answer:
0.45 g
Explanation:
Step 1: Given data
- Molar mass of methionine (M): 149.21 g/mol
- Volume of the solution (V): 20 mL
- Concentration of the solution (C): 150 mM
Step 2: Calculate the moles of methionine (n)
We will use the following expression.
n = C × V
n = 150 × 10⁻³ mol/L × 20 × 10⁻³ L
n = 3.0 × 10⁻³ mol
Step 3: Calculate the mass of methionine (m)
We will use the following expression.
m = n × M
m = 3.0 × 10⁻³ mol × 149.21 g/mol
m = 0.45 g
