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3 years ago

Which condition in a nebula would prevent nuclear fusion

Chemistry

1 answer:

White raven [17]3 years ago

3 0

this is not my work

-Brooks Nelson

Brooks Nelson, Chemist at University of Florida

Answered Oct 12, 2018 · Author has 368 answers and 54.1k answer views

My limited understanding is you need pressure, temperature and enough elements that can fuse. If the temperature and pressure aren't high enough and/or you don't have enough elements that can fuse, then no fusion.

In fact I've never heard of fusion in a nebula, only in a star. The exception being a brown dwarf, which is considered substellar at 10 to 90 Jupiters in mass, and they can fuse deuterium (if over 13J) and also lithium (if over 60 J). But the burn through all of it in about 10 million years and wouldn't emit light like a main sequence star would.

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While performing an experiment involving a reaction between two chemicals, the scientist observes that the reaction container ha

Mama L [17]

If he measures "How much warmer [which he should]," then he would be measuring quantity and B would be the answer.

Since he is only observing that the reaction container has become warmer, then qualitative data is all that he is measuring. The answer is A

C is not true "How much warmer" is not answered.

The observation is not a number but it is a property. It is qualitative as color would be. (If a reaction changes color it would be qualitative).

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3 years ago

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What is the name of the highest temperature at which it is possible to liquefy a gas with any amount of pressure?

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The answer is critical point

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3 years ago

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above relation, we get:

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M