Answer:
+15.8°
Explanation:
The formula for the observed rotation (α) of an optically active sample is
α = [α]<em>lc
</em>
where
<em>l</em> = the cell path length in decimetres
<em>c</em> = the concentration in units of g/100 mL
[α] = the specific rotation in degrees
1. Convert the concentration to units of g/100 mL
2. Calculate the observed rotation
Answer:
Buffalo and Watertown.
Explanation:
This is correct as the Interior lowlands are located ib the US. These are correct as the others are in the US but aren't located anywhere near the Interior Lowlands.
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L
There is no way to know which reaction requires which catalyst. However, if you apply copper to a reaction where it does act as a catalyst, the rate of reaction will be much faster as it lowers the activation energy for successful collisions.
Answer:
4.66 x 10^8 yr
Explanation:
The age of the rock can be calculated using the equation:
ln (N/N₀) = - kt where N is the quantiy of radioisotope decayed and N₀ is the initially quantity present of the radioisotope; k is the decay constant, and t is the time.
Now from the data , we have 78 argon-40 atoms for every 22 potassium-40 atoms, we can deduce that originally we had 22 + 78 = 100 atoms of potassium-40 so this is our N₀.
When we look at the equation, we see that k is unknown, but we can calculate it from the half-life which is given by the equation:
k = 0.693/ t half-life = 0.693/ 1.3 x 10⁹ yr = 5.33 x 10⁻¹⁰ yr⁻¹
Now we are in position to answer the question.
ln ( 78/100 ) = - (5.33 x 10⁻¹⁰ yr⁻¹ ) t
- 0.249 = - 5.33 x 10⁻¹⁰ yr⁻¹ t
0.249/ 5.33 x 10⁻¹⁰ yr⁻¹ = t
4.66 x 10^8 yr
