All categories

3 years ago

In an exothermic reaction, energy is

2 answers:

7 0

Answer: Exothermic reactions are reactions or processes that release energy, usually in the form of heat or light. In an exothermic reaction, energy is released because the total energy of the products is less than the total energy of the reactants.

Explanation:

Nutka1998 [239]3 years ago

5 0

In Exothermic reaction energy is Thrown/Relased to the surroundings.
In endothermic reaction, energy is absorbed from the surroundings.

Trick to identify:-

Exothermic reaction has +∆H sign on product side
Endothermic reaction has -∆H sign on product side or +∆H sign on reactant side

You might be interested in

How to tell if something is an electrolyte

schepotkina [342]

Answer: If it has ions, it is an electrolyte

Explanation:

Let's start by explaining that electrolytes are compounds that contain charged particles or<u> ions</u>, which can be cations (positive ions) or anions (negative ions).

So, it is this composition that makes an electrolytic material conduct electricity.

In this sense, the way to identify if a material is an electrolyte or not, is knowing whether it is composed of ions or not.

8 0

3 years ago

All fires require oxygen () There is no oxygen in that room (T) .

Marina CMI [18]

The sentences are invalid and unsound.

<h3><u>Explanation</u>:</h3>

The fire is defined as the vigorous oxidation of a substance. Now oxidation can occur in presence of any oxidising agent. Like magnesium in presence of nitrogen in high temperature with a dazzling brownish flame to produce magnesium nitride. So fire can be produced in absence of oxygen.

Oxygen is present everywhere in world. So production of a whole room without oxygen is very tough to produce and costly process. So its very unsound.

4 0

3 years ago

What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each

suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
A = area of plates and
d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

7 0

2 years ago

the imagine above shows to opposite forces acting on a rolling cart, what can we say is true about affect of the forces on the c

cluponka [151]

Answer:

Here is my answer...

Explanation:

The cart will connect with the opposite force, and then the cart will come to a shuddering stop before moving in the direction of the oposite force.

Hope I helped! :)

7 0

2 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

3 years ago