Which diastereomer of oct−4−ene yields a single meso compound, (4r, 5s)−4,5−dibromooctane on reaction with br2?

In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)

Brrunno [24]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.6 L

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}$

${V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L$

${V_2}=0.27\ L$

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

8 0

2 years ago

A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 3

Sati [7]

Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C

(58 g . 0.100) / 0.580 = 10 g of H

48 g of C / 12 g/mol = 4 mol

10 g of H / 1g/mol = 10 moles

7 0

2 years ago

2.1. The lithium ion in a 250,00 mL sample of mineral water was

juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

Moles Mg²⁺:

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

Moles B(C₆H₄)₄ = Moles Lithium:

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

4 0

2 years ago

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K

Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

$\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S$

Thus, by combining them, we obtain:

$-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}$

Which is related to the general line equation:

$y=mx+b$

Whereas:

$y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}$

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

8 0

2 years ago

40pionts

Pachacha [2.7K]

Problem One (left)

This is just a straight mc deltaT question

Givens

m = 535 grams

c = 0.486 J/gm

tf = 50

ti = 1230

Formula

E = m * c * (ti - tf)

Solution

E = 535 * 0.486 * ( 1230 - 50)

E = 535 * 0.486 * (1180)

E = 301077

Answer: A

Problem Two

This one just requires that you multiply the two numbers together and cut it down to 3 sig digits.

E = H m

H = 2257 J/gram

m = 11.2 grams

E = 2257 * 11.2

E = 25278 to three digits is 25300 Joules. Anyway it is the last one.

Three

D and E are both incorrect for the same reason. The sun and stars don't contain an awful lot of Uranium (1 part of a trillion hydrogen atoms). It's too rare. The other answers can all be eliminated because U 235 is pretty stable in its natural state. It has a high activation complex.

Your best chance would be enriched Uranium (which is another way of saying refined uranium). That would be the right environment. Atomic weapons and nuclear power plants (most) used enriched Uranium. You can google "Little Boy" if you want to know more.

Answer: B

Four

The best way to think about this question is just to get the answer. Answer C.

A: incorrect. Anything sticking together implies a larger and larger result. Gases don't work that way. They move about randomly.

B: Wrong. Heat and Temperature especially depend on movement. Stopping is not permitted. If a substance's molecules stopped, the substance would experience an extremely uncomfortable temperature drop.

C: is correct because the molecules neither stop nor do they stick. The hit and move on.

D: Wrong. An ax splitting something? That is not what happens normally and not with ordinary gases. It takes more energy that mere collisions or normal temperatures would provide to get a gas to split apart.

E: Wrong. Same sort of comment as D. Splitting is not the way these things work. They bounce away as in C.

Five

Half life number 1 would leave 0.5 grams behind.

Half life number 2 would leave 1/2 of 1/2 or 1/4 of the number of grams left.

Answer: 0.25

Answer C

6 0

2 years ago