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Ugo [173]
2 years ago
12

Use complete sentences to differentiate between acids and bases on the basis of their behavior when dissolved in water Give an e

xample of each type.
Physics
1 answer:
grigory [225]2 years ago
4 0

Explanation:

An acid is a substance that interacts with water to produce excess hydroxonium ions, H₃O⁺ in an aqueous solution.

The hydroxonium ions formed by the chemical bonding between the oxygen of water molecules and the protons released by the acid due to its excess ionization.

  For example:

              HCl          +         H₂O ⇄       H₃O⁺         +             Cl⁻

               acid                    water         hydroxonium         chloride

A base is a substance that interacts with water to yield excess hydroxide ions, OH⁻ in an aqueous solutions:

            NH₃        +          H₂O       ⇄       NH₄⁺     +            OH⁻

        Ammonia            water          ammonium              hydroxide

These explanations are based on the arrhenius theory.

learn more:

Bronsted-lowry theory  brainly.com/question/4083753

Acid reaction brainly.com/question/5273689

#learnwithBrainly

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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
2 years ago
a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the
ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: F=m*a with this we can get both accelerations; solving for acceleration a=\frac{F}{m}. Now a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2}) anda_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2}). Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, x=\frac{1}{2}*a_{girl}*t^{2} and15.0-x=\frac{1}{2}*a_{sled}*t^{2}, solving for the time we get:t=\sqrt{\frac{2x}{a_{girl} } } and t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } } with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get:\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }. Finally we get:\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} } and replacing the values we have got:\frac{x}{0.14} =\frac{(15.0-x)}{0.73} so 5.33*x=15-x so x=2.37 (meters).

5 0
3 years ago
PLEASE HELP ME ASAP PLEASEEEEE
AleksAgata [21]
I believe its the law of inertia
6 0
3 years ago
If two forces are in the same direction then do they cancel each other out
stellarik [79]
Im pretty sure that they do

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3 years ago
PLEASE HELP ME WITH THSES QUESTIONS 3 SENTENCES PLEASE I BEG YOU!
JulsSmile [24]
Question 2 is because the passengers have inertia, which is a tendency to resist change in motion
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