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tangare [24]
3 years ago
15

How could an exothermic reaction be identified in the lab?

Chemistry
1 answer:
Lelu [443]3 years ago
5 0

<em>The </em><em>reaction </em><em>in </em><em>which </em><em>the </em><em>heat </em><em>is </em><em>released </em><em>to </em><em>the </em><em>surrounding </em><em>is </em><em>called </em><em>exothermic </em><em>reaction</em><em>.</em>

<em>And </em><em>exothermic </em><em>reaction </em><em>can </em><em>be </em><em>identified </em><em>while </em><em>doing </em><em>chemical </em><em>reactions</em><em>.</em><em>,</em><em> </em><em>when </em><em>heat </em><em>is </em><em>released</em><em> </em><em>then </em><em>it </em><em>is </em><em>exothermic </em><em>reaction</em><em>. </em>

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Consider an electron with charge −e−e and mass mmm orbiting in a circle around a hydrogen nucleus (a single proton) with cha
PtichkaEL [24]

Answer:

Explanation:

The net force on electron is electrostatic force between electron and proton in the nucleus .

Fc = \frac{1}{4\pi\epsilon} \times \frac{e\times e}{r^2}

This provides the centripetal force for the circular path of electron around the nucleus .

Centripetal force required = \frac{m\times v^2}{r}

So

\frac{m\times v^2}{r}=\frac{1}{4\pi\epsilon} \times \frac{e\times e}{r^2}

v^2=\frac{e^2}{4\pi \epsilon m r}

v=(\frac{e^2}{4\pi \epsilon m r})^{\frac{1}{2} }

5 0
3 years ago
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ipn [44]

Answer:

A 400,000 kg object and a 100,000,000kg object

Explanation:

The objects with the most mass between them will have the greatest gravitational attraction.

This is why the last option is the right choice.

The reason for this is based on the Newton's law of universal gravitation which states that:

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So, the more the mass, the greater the gravitational attraction between two bodies.

4 0
3 years ago
The energy in eV for light with a wavelength of 6250 angstroms is _. Note - there are 1.6 x 10-12 erg in 1 eV.
vivado [14]

Answer:

2 eV

Explanation:

The energy of a photon of light is given by the formula

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength of the photon

In this problem we have:

h=6.63\cdot 10^{-34} Js

c=3.0\cdot 10^8 m/s

\lambda=6250 A = 6250\cdot 10^{-10} m is the wavelength of the photon

Therefore, the energy in Joules is

E=\frac{(6.63\cdot 10^{-34})(3.0\cdot 10^8)}{6250\cdot 10^{-10}}=3.2\cdot 10^{-19}J

We want to convert this energy into electronvolts: we know that the conversion factor is

1 eV = 1.6\cdot 10^{-19}J

Therefore,

E=\frac{3.2\cdot 10^{-19}}{1.6\cdot 10^{-19}}=2 eV

5 0
3 years ago
How many stereoisomers could exist for the following structure?<br><br> 4<br> 8<br> 16<br> 32
8_murik_8 [283]

Answer:

64

Explanation:

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3 0
2 years ago
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Answer:

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Explanation:

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