Answer:
5.2 x 10⁻⁴ M.
Explanation:
- The relationship between gas pressure and the concentration of dissolved gas is given by Henry’s law:
<em>P = kC</em>
where P is the partial pressure of the gaseous solute above the solution.
k is a constant (Henry’s constant).
C is the concentration of the dissolved gas.
- At two different pressures, there is two different concentrations of dissolved gases and is expressed in a relation as:
<em>P₁C₂ = P₂C₁,</em>
P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.
P₂ = 0.76 atm, C₂ = ??? mol/L.
<em>∴ C₂ = (P₂C₁)/P₁ =</em> (0.76 atm)(6.8 x 10⁻⁴ mol/L)/(1.0 atm) = <em>5.168 x 10⁻⁴ mol/L ≅ 5.2 x 10⁻⁴ M.</em>
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
<h3>
Answer:</h3>
56.11 g/mol
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Compound] KOH
<u>Step 2: Identify</u>
[PT] Molar Mass of K - 39.10 g/mol
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mass of H - 1.01 g/mol
<u>Step 3: Find</u>
39.10 + 16.00 + 1.01 = 56.11 g/mol
Take a hypothetical sample of exactly 100 grams of the solution.
(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea
((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O
(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea
Answer:
i think i had the same question as you, I put "A car driving on a straight freeway at 60 miles per hour" which was correct for me.
Explanation:
