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2 years ago

What occurs during an exothermic change

2 answers:

6 0

In an exothermic reaction, there is a transfer of energy to the surroundings in the form of heat energy. The surroundings of the reaction will experience an increase in temperature. Many types of chemical reactions are exothermic, including combustion reactions, respiration & neutralization reactions of bases & acids.

nikklg [1K]2 years ago

3 0

The system releases energy to its surroundings.

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A scientist wants to learn about Earth's oceans. Which of the following

ahrayia [7]

Hydrosphere - all the waters on the earth’s surface.

8 0

2 years ago

A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4m .determine the acceleration of the bike.

sergeinik [125]

Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s
Find: Acceleration
Analysis:
Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)
(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)
50.41 m/s²=(70.8 m)a
a=0.712 m/s²

5 0

2 years ago

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

Readme [11.4K]

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
substituting the coordinates of the two charges, we get
$d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m$

2) Then, we can calculate the electrostatic force between the two charges $q_1$ and $q_2$ , which is given by
$F=k_e \frac{q_1 q_2}{d^2}$
where $k_e=8.99\cdot10^{9} Nm^2C^{-2}$ is the Coulomb's constant.
Substituting numbers, we get
$F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N$
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.

7 0

3 years ago

Caleb rides his horse with a constant speed of 12 miles per hour. How far can she travel in 3 1/2 hours?

baherus [9]

12 × 3= 36
12 × 1/2 = 6
36+ 6= 42

6 0

3 years ago

A satellite is launched to orbit the Earth at an altitude of 3.25 107 m for use in the Global Positioning System (GPS). Take the

Korolek [52]

Answer:Orbital period =21.22hrs

Explanation:

given that

mass of earth M = 5.97 x 10^24 kg

radius of a satellite's orbit, R= earth's radius + height of the satellite

6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m

Speed of satellite, v= $\sqrt GM/R$

where G = 6.673 x 10-11 N m2/kg2

V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)

V =10,241082.2

v= 3,200.2m/s

a) Orbital period

$\sqrt GM/R$ = $\frac{2\pi r}{T}$

V= $\frac{2\pi r}{T}$

T= 2 $\pi$ r/ V

= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s

=76,385.1 s

60 sec= 1min

60mins = 1hr

76,385.1s =hr

76,385.1/3600=21.22hrs

3 0

3 years ago