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3 years ago

If approximately equal volumes of an acid and a bace of a equal strength are combined the resulting solutin should have a pH of

Physics

2 answers:

marusya05 [52]3 years ago

4 0

Answer : The correct option is, around 7

Explanation :

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

$pH=-\log [H^+]$

The pH range is, 1 to 14.

When the value of pH is less then 7 then the solution will be acidic.

When the value of pH is more then 7 then the solution will be basic.

When the value of pH is equal to 7 then the solution will be neutral.

As per question, when approximately equal volumes of an acid and a base of a equal strength are combined then the resulting solution should have a pH of around 7 because when equal amount of acid and base are combined it form neutral solution and the neutral solution has pH around 7 or equal to 7.

Hence, the correct option is, around 7

Phoenix [80]3 years ago

3 0

The answer is around 7.

As the acid and the base are of equal strength, they neutralize each other and the resulting solution is neutral (pH≈7).

Hope this helps you!

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Which law states that absolute zero cannot be reached?

padilas [110]

The third law of thermodynamics,the principle of temperature.

This law states that the entropy at 0 is always equel to 0.
This means that it is impossible to cool down a perfect 0 or absolute 0(-273.15 C)

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2 years ago

Read 2 more answers

Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T

Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

$T-f-T'=m_Ba$

Where $m_B$ =Mass of block B

Substitute the values

$50-10-T'=m_B\times 0=0$

$T'==40 N$

For block A

$T'-f'=m_Aa$

Where $m_A=$ Mass of block A

Substitute the values

$40-f'=m_A\times 0=0$

$f'=40 N$

Hence, the friction force acting on block A=40 N

3 0

2 years ago

A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic

Ilia_Sergeevich [38]

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$ :

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$

4 0

2 years ago

Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of

FrozenT [24]

Gravity obeys the inverse square law. At 6400 km above the center of the Earth (Earth's surface) you weigh x. Twice that reduces your weight to 1/4th. Four times that height reduces your weight to 1/16th. 4 times 6400 km is 25,600 km. But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer: 19,200 km.

Incidentally, it doesn't exactly work the opposite way. At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.

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2 years ago

I WILL MARK YOU THE BRAINLIEST NO LINKS

nasty-shy [4]

yeah it definitely 2 .:) :)

3 0

2 years ago