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3 years ago

What are the products of a neutralization reaction

Physics

2 answers:

Westkost [7]3 years ago

6 0

Answer: The correct answer is- Salt and water.

Neutralization is a type of chemical reaction, which involves a strong acid and strong base as reactants in order to form salt and water as product. In other words, when strong acid is allowed to react with strong base, it produces salt and water as a product and this is called as neutralization reaction.

For example- HCl + NaOH → NaCl + H2O

(acid) (base) (salt) (water)

Thus, salt and water is the right answer.

Lapatulllka [165]3 years ago

4 0

There are no other products of neutralization reactions other than salt and water. (Salt is a much more broader term than just table salt by the way.)

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3 years ago

Express the vector R B in terms of A, B, C, and Ď, the edges of a parallelogram.

Vadim26 [7]

Answer:

R=0.5B+0.5C+2A+D

Explanation:

By the triangular law of vector addition

vector R= vector B- vector D

As A,B,C,D are edges of the parallelogram,

A is parallel to D but opposite in direction.

Therefore

$A = (-D)$ ; $A//D$ ;

$2A=-2D$

B is parallel to C and in same direction.

$B//C\\B=C\\$

$0.5B=0.5C\\$

$R= B-D;\\R= 0.5B+0.5B-D;\\R=0.5B+0.5C-D;\\R=0.5B+0.5C-2D+D;\\R=0.5B+0.5C+2A+D;$

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2 years ago

The binding of acetylcholine to its receptor at the neuromuscular junction causes the opening of a

Rufina [12.5K]

Answer: opening of the nicotinic acetylcholine receptor channels.

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2 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.