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lapo4ka [179]
3 years ago
7

Write a balanced chemical equation for the decomposition of solid sodium sulfide in aqueous solution.

Physics
2 answers:
Ganezh [65]3 years ago
7 0

Answer:

Na2S(s) → 2Na+(aq) + S2-(aq) is the only one that balances the reactants and the products.

Explanation:

lilavasa [31]3 years ago
3 0
The answer is Na2(s)→2Na+(aq)+S²-(aq)
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Dois carros, a e b, móveis se em uma estrada retilínea com volocidade constante,vª =20m/s e v=18 m/s , respectivamente. O carro
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3 years ago
True or False. Recent findings lend strong support to the theory that a black hole lies at the center of the Milky Way and of ma
Genrish500 [490]

Answer:

True

Explanation:

In 2019 was taken the first image of a supermassive black hole in the center of the galaxy M87 with the Event Horizon Telescope, which is a network of radio telescope located in different points of the Earth, with the purpose of making a telescope of the size of the Earth.

A radio telescope is an antenna that is capable to perceive the light in the radio part of the electromagnetic spectrum¹.

It is important to notice that in the picture what it can be seen is the effect that the black hole has in the nearby stars.

Key terms:

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7 0
3 years ago
The vaule of the g constant (the acceleration of all objects due to gravity)on earth is
True [87]
 The answer is:  " \frac{9.8 m}{s^2}" .
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6 0
3 years ago
A rope is tied to a box and used to pull the box 2.3 m along a horizontal floor. The rope makes an angle of 30∘ with the horizon
Savatey [412]

Answer:

Answered

Explanation:

A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.

W= FS cosθ

θ= 90 ⇒cos90 = 0 ⇒W= 0

B) work done by tension

W= Tcosθ×S= 5cos30×2.30= 10J

C) Work done by friction force

W= f×s=1×2.30= 2.30 J

D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.

E) The net work done= Work done by tension in the rope - frictional work

=10-2.30= 7.7 J

6 0
2 years ago
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
Lubov Fominskaja [6]

Answer:

The moment of inertia is I=\frac{M}{12} a^{2}

Explanation:

The moment of inertia is equal:

I=\int\limits^a_b {r^{2} } \, dm

If r is -\frac{a}{2}

and dm=\frac{M}{a} dr

I=\int\limits^a_b {r^{2}\frac{M}{a}  } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}

I=\frac{M}{a} \int\limits^a_b {r^{2}  } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\  I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}

7 0
3 years ago
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