Answer:
5.12x10¹¹ millimeters
Explanation:
Milli is a prefix used in science and engineering to decribe the number as the exponent x10⁻³. In the prefix kilo, the number is at the exponent x10³.
5.12x10⁵ kilometers are:
5.12x10⁵ kilometers * (1000m / 1km) = 5.12x10⁸ meters
5.12x10² meters * (1m / 1000millimeters) = 5.12x10¹¹ millimeters
1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.
2) So, one alkyl group is CH3 and second one can be CH or CH2.
3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2.
4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.
The ratio remains the same, 3:2 ie 6:4
Answer:
25.6g de HF son producidos
Explanation:
<em>...¿Cuánto HF es producido?</em>
Para resolver este problema debemos convertir la masa de cada reactivo a moles usando su masa molar. Como la reacción es 1:1, el reactivo con menor número de moles es el reactivo limitante. Con las moles del reactivo limitante podemos obtener las moles de HF y su masa así:
<em>Moles CaF2:</em>
Masa molar:
1Ca = 40g/mol
2F = 19*2 = 38g/mol
40+38 = 78g/mol
50g CaF2 * (1mol/78g) = 0.641 moles CaF2
<em>Moles H2SO4:</em>
Masa molar:
2H = 2g/mol
1S = 32g/mol
4O = 64g/mol
98g/mol
100g H2SO4 * (1mol / 98g) = 1.02 moles H2SO4
Como las moles de CaF2 < Moles H2SO4: CaF2 es reactivo limitante.
<em>Moles HF usando la reacción:</em>
0.641 moles CaF2 * (2mol HF / 1mol CaF2) = 1.282 moles HF
<em>Masa HF:</em>
Masa molar:
1g/mol + 19g/mol = 20g/mol
1.282 moles HF * (20g/mol) =
<h3>25.6g de HF son producidos</h3>
Henry's law constant for oxygen is 0,0013 mol/L·<span>atm. Air has 21,0% oxygen.
concentration of oxygen at 1 atm: 0,0013 mol/L</span>·atm · 0,21 · 1 atm = 0,000273 mol/l.
concentration of oxygen at 1 atm: 0,0013 mol/L·atm · 0,21 · 0,892 atm = 0,000243 mol/l.
difference in concentration: 0,000273 - 0,000243 = 0,00003 mol/L.
n(oxygen) = 0,00003 mol/L · 4,40 L = 0,000132 mol.