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olganol [36]
4 years ago
13

Which substance has the lowest [H+] concentration but is still considered an acid?

Chemistry
2 answers:
BabaBlast [244]4 years ago
8 0
<h3>Answer:</h3>

                  Milk

<h3>Explanation:</h3>

                     The pH of all given substances is as follow;

                                       Milk  =  6.5 to 6.7

                                       Blood  =  7.35

                                       Gastric Fluid  =  1.5 to 3.5

                                       Household Lye  =  13.5

Therefore, we can exclude Blood and Household Lye because there pH is greater than 7 hence, they are either neutral (Blood) or Basic (Household Lye) in nature. Now, we are left with Milk and Gastric Fluids.

As we know greater the concentration of H⁺ ions, smaller is the pH hence, gastric fluid having less pH than Milk will contain more H⁺ ions therefore, more acidic than Milk. While, Milk containing less H⁺ ions (greater pH than Gastric fluid) is less acidic and contains less number of H⁺ ions.

attashe74 [19]4 years ago
6 0

the correct answer is a, milk


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Why can a solution be classified as a mixture?
Lapatulllka [165]

Answer:

A. It is made up of only one substance

Explanation:

6 0
3 years ago
What is the pH if 1mL of 0.1M HCl is added to 99mL of pure water?
coldgirl [10]

Answer:

pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,2

Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:

7,0 = 7,2 + log₁₀ \frac{[HPO4^{2-}] }{[H2PO4^{-}]}

Ratio obtained is:

0,63 = \frac{[HPO4^{2-}] }{[H2PO4^{-}]}

As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M

As the amount added of HCl is 0,001 M the concentrations in equilibrium are:

H₂PO₄⁻   ⇄   HPO4²⁻ +        H⁺

0,1 M +x      0,063M -x  0,001M -x -<em>because the addition of H⁺ displaces the equilibrium to the left-</em>

Knowing the equation of equilibrium is:

K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}

Replacing:

6,20x10⁻⁸ = \frac{[0,063-x][0,001-x]}{[0,1+x]}

You will obtain:

x² -0,064 x + 6,29938x10⁻⁵ = 0

Thus:

x = 0,063 → No physical sense

x = 0,00099990

Thus, [H⁺] in equilibrium is:

0,001 M - 0,00099990 = 1x10⁻⁷

Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =

-log₁₀ [1x10⁻⁷] = 7,0

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!

I hope it helps!

5 0
3 years ago
PLS HELP
serious [3.7K]

4. True

5. The amplitude of the ripples decreases as the circumference of the circle increases.

<h3>What is a wave?</h3>

A wave is a disturbance that moves energy from one place to another.

Radio waves, gamma-rays, visible light, and all the other parts of the electromagnetic spectrum are electromagnetic radiation. Electromagnetic radiation can be described in terms of a stream of mass-less particles, called photons, each travelling in a wave-like pattern at the speed of light. Hence, the statement is true.

As the disturbance moves outwards the energy it carries is spread over a larger and larger region called the wavefront. For example, the ripples in the water lie in a circle with an ever-increasing circumference. The amplitude of the ripples decreases as the circumference of the circle increases.

Learn more about the Electromagnetic wave here:

brainly.com/question/3101711

#SPJ1

6 0
2 years ago
Anybody know how to do this? Max points
svlad2 [7]
Deleting it and downloading it back again
5 0
3 years ago
A system does 125 J of work and cools down by releasing 438 J of heat. The change in internal energy is ____________ J.
Jet001 [13]

Given that,

Work done by the system = 125 J

Energy released when it cools down = 438 J

To find,

The change in internal energy.

Solution,

As heat is released by the system, Q = -438 J

Work done by the system, W = -125 J

Using the first law of thermodynamics. The change in internal energy is given by :

\Delta U=Q-W\\\\=-438-125\\\\=-563\ J

So, the change in internal energy is 563 J.

8 0
3 years ago
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