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gulaghasi [49]
3 years ago
13

What changes in space result from the death of massive stars?

Chemistry
1 answer:
lana [24]3 years ago
6 0

Whenever the fuel is being used up, a star explodes and the energy leakage from a star's core ceases.

Explanation:

The dying star expands in the "Red Giant," before even the inevitable collapse starts, due to nuclear reactions just outside of the core.

It becomes a white dwarf star when the star has almost the same density as the Sun. If it's much larger, a supernova explosion could take place and leave a neutron star away. However, if it is very large–at least three times the Sun's mass–the crumbling core of the star, nothing will ever stop it from crumbling. The star is imploding into a black hole, an endless gravitational loop in space.

You might be interested in
Is ice melting in a drink a chemical change?
konstantin123 [22]
Nope, it's a physical change. When ice melts it changes into liquid water, it's still water. Changes in which no new substances are formed are physical changes, hence this is a physical change.
6 0
3 years ago
Sodium sulfide reacts with hydrochloric acid to produce hydrosulfuric acid and sodium chloride. (you need to write and balance t
VikaD [51]

Answer:

2.5 moles of NaCl

Explanation:

The balanced chemical reaction equation is shown in the image. Since it takes 2 moles of Hydrochloric acid to form two moles of sodium. Chloride, then 2.5 moles of hydrochloric acid should also form 2.5 moles of sodium chloride according to the balanced reaction equation.

4 0
3 years ago
Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?
charle [14.2K]

Answer:

The answer is "41.23 \ L\  N_2"

Explanation:

2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049  \frac{\ g NO_3^{-}}{mol})} \times  \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\

=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23

8 0
3 years ago
What particle decay is this? 210 83 Bi→210 84 Po ​
sesenic [268]

Answer:

Beta emission

Explanation:

In beta emission, a neutron is converted into a proton thereby emitting an electron and a neutrino. A neutrino is a particle that serves to balance the spins.

When a nucleus undergoes beta emission, the mass number of the parent and daughter nuclei remain the same while the atomic number of the daughter nucleus is greater than that of its parent by one unit.

Hence, in beta emission, the daughter nucleus is found one pace to the right of the parent in the periodic table.

3 0
3 years ago
You are making a 1L solution that you want to be buffered with ammonia and ammonium. However ammonia is a gas that is difficult
zloy xaker [14]

Answer:

54 grams ammonium chloride and 40 grams sodium hydroxide

Explanation:

A buffer is a solution that contains either a weak acid and its salt or a weak base and its salt, the solution is resistant to changes in pH. This means that, a buffer is an aqueous solution of either a weak acid and its conjugate base or a weak base and its conjugate acid.

A Buffer is used to maintain a stable pH in a solution, buffers can neutralize small quantities of additional acid of base. For any buffer solution, there is always a working pH range and a set amount of acid or base that can be neutralized before the pH will change. The amount of acid or base that can be added to a buffer before changing its pH is called its buffer capacity.

A good buffer mixture is supposed to have about equal concentrations of its both components. It is a rule of thumb therefore, that a buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other component.

The implication of this is that the ammonium chloride and sodium hydroxide should be of approximately the same concentration. If the masses are dissolved as shown in the answer, then we will have 1molL-1 of each component of the buffer in accordance with the rule of thumb stated above.

4 0
3 years ago
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