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EastWind [94]
3 years ago
7

In the evolution of extremely large stars, they begin as supergiants or supernovas, if the star nucleus is large enough it will

continue to contract and form neutron stars, and finally, if the contractions continue, everything around the neutron star and the neutron star itself will be pulled into what is called a
Chemistry
2 answers:
fenix001 [56]3 years ago
6 0
Black hole, or a singularity contained within an event horizon through which no light can escape.
LiRa [457]3 years ago
5 0

Answer: The answer would be black hole

Explanation: if you’re using Edgenty they explain it in the lesson

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Why should you pay attention to the physical states of reactants and products when writing elquilibrium constant expressions?
babunello [35]
Answer is: because pure liquids (<span>shown in </span>chemical reactions<span> by appending (</span>l)<span> to the </span>chemical formula) and solids (<span>shown in </span>chemical equations by appending (s)<span> to the </span>chemical formula) not go in to he equilibrium constant expression, only gas state (shown in chemical reactions by appending (g) to the chemical formula) reactants and products go in to he equilibrium constant expression.
For example, equilibrium constant expression Kp for reaction:
A(s) + 2B(s) ⇄ 4C(g) + D(g).<span>
will be: Kp = [C]</span>⁴<span>·[D].
But for reaction </span>A(g) + 2B(g) ⇄ 4C(g) + D(g), will be:<span>
Kp = [C]</span>⁴<span>·[D] / [A]·[B]².</span>
3 0
4 years ago
A teacher reacts 84.0 g of sodium bicarbonate with excess hydrochloric acid. What is the percent yield for sodium chloride if 56
Bond [772]

Answer:

b

Explanation:

5 0
3 years ago
Pls that is the picture ​
Fiesta28 [93]
Do what it says easy
3 0
3 years ago
The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilo
Lynna [10]

Answer:

NA = 6.8 E-12 Kg H2(g) / hour

Explanation:

steady-state diffusion of A through non-diffuser B:

  • NA = (DAB/RTz)(p*A1 - p*A2)

∴ (A): H2(g)    

∴ (B): Pd

∴ DAB = 1.7 E-8 m²/s

∴ p*A1 = 2.0 Kg H2 / m³ Pd

∴ p*A2 = 0.4 Kg H2 / m³ Pd

∴ z = 6 mm = 6 E-3 m

∴ T = 600°C ≅ 873 K

∴ R = 8.314 J/mol.K = 8.314 N.m/mol.K

⇒ NA = ((1.7 E-8)/(8.314)(873)(6 E-3))(2.0 - 0.4)

⇒ NA = 6.246 E-10 mol/s.m³

for A = 0.25 m²

⇒ volume (v) = A×z = (0.25)(6 E-3) = 1.5 E-3 m³

∴ Mw H2(g) = 2.016 g/mol

⇒ NA = (6.246 E-10 mol/s.m³)(1.5 E-3 m³)(2.016 g/mol)(Kg/1000 g)(3600 s/h)

⇒ NA = 6.8 E-12 Kg H2(g)/h

4 0
3 years ago
Someone help me with this. Imagine you drank a big glass of water today. Ten years later,if you go to Antarctica and drink a gla
Artyom0805 [142]
Well when you would "get rid of" the water it would eventually be led to some ocean, pond, lake, etc. There it would eventually evaporate, rising up to the sky. This is where it would move around with clouds until it rained again. 
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3 years ago
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