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3 years ago

Substances that prevent certain chemical reactions are called ___.

Physics

1 answer:

yuradex [85]3 years ago

5 0

Substances that prevent or reduce the rate of a chemical reaction are called as inhibitors.

Explanation:

Inhibitors are those substances that either prevent or reduce the rate of a chemical reaction. Ex: Acetanilide slows down the decomposition of hydrogen peroxide solution.

Inhibitors even reduce the effectiveness of catalysts without hindering their composition. There are primarily two kinds of inhibitors- reversible and irreversible. Where, reversible inhibitors slows the rate of any reaction, irreversible inhibitors completely prevents the unwanted reactions.

Where catalysts are used to increase the rate of a reaction, inhibitors stand as a wall for the unwanted reactions to occur. For example, in case of silver tarnishing, to prevent silver jewellery being tarnished, a thin layer of rhodium is applied on it which works like an inhibitor for formation reaction of silver sulphides (the tarnish).

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Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0

3 years ago

xperiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are gi

creativ13 [48]

Answer:

a) b) d)

Explanation:

The question is incomplete. The Complete question might be

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.

a)2 N; 2 N

b) 200 N; 200 N

c) 200 N; 201 N

d) 2 N; 2 N; 4 N

e) 2 N; 2 N; 2 N

f) 2 N; 2 N; 3 N

g) 2 N; 2 N; 5 N

h ) 200 N; 200 N; 5 N

For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces

a) 2+(-2)=0 here minus sign is to show the opposing firection of force

b) 200+(-200)=0

c) 200+(-201) $\neq$ 0

d) 2+2+(-4)=0

e) 2+2+(-2) $\neq$ 0

f) 2+2+(-3) $\neq$ 0; 2+(-2)+3 $\neq$ 0

g) 2+2+(-5) $\neq$ 0; 2+(-2)+5 $\neq$ 0

h)200 + 200 +(-5) $\neq$ 0; 200+(-200)+5 $\neq$ 0

6 0

3 years ago

The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag

allochka39001 [22]

In order to get the propoerty of work you need to use the following formula
work = force times distance
replacing data you will get:
W = (1.500) (3)
W = 4.500 NM
The answer should be in NM. So it will be 4500 NM againts the force of gravity

4 0

3 years ago

A total electric charge of 4.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm. The pot

nevsk [136]

for what we can get from our side then we will see if they are being made in a letter or something which is leading enormous growth in cities with brick and its own way that it will not on the market and its not a good life for us to stay in and 

7 0

1 year ago

En un recipiente de cobre de forma rectangular de aproximadamente 3.5 m largo por 4.5m ancho tiene una temperatura en temporada

Darina [25.2K]

Answer:

A_f= 15,769 m²

Explanation:

Este es un ejercicio de dilatación térmica,

ΔA = (2α) A₀ ΔT

el arrea de recipiente

A₀ = L A

A₀ = 3,5 4,5

A₀= 15,75 m²

el coeficiente de dilatación térmica es alfa = 16,6 10⁻⁶ C⁻¹

calculemos

ΔA = 2 16,6 10⁻⁶ 15,75 ( 40 -4)

ΔA = 522,9 (36) 10⁻⁶

A= 1,88 10⁻² m2

el cambio de volumen es

ΔA = A_f – A₀

A_f = A₀ + ΔA

A_f= 15,75 +1,88 10⁻²

A_f= 15,769 m²

3 0

3 years ago