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FromTheMoon [43]
2 years ago
12

an acid accepts H and removes them from a solution, is a solution where the water molecules are intact

Chemistry
1 answer:
n200080 [17]2 years ago
8 0
<h3>An acid accepts H and removes them from a solution</h3>

When placed in water, acids, bases, and salts dissociate (separate) into electrolytes (ions). Salts dissociate into a cation (that is not H+) and an anion (that is not OH-), whereas acids and bases dissociate into H+ and an anion. An acid separates into anions and hydrogen ions (H+). Strong acids produce a high concentration of H+ by dissociating every single one of their molecules . Water-based solutions,

Acid:

When a material or chemical is in solution, it releases hydrogen ions (H+), which are known as acids. All hydrogen ions (H+) and chloride ions (Cl-), which are normally bound together by ionic bonding, dissociate (separate) in water when exposed to a strong acid like hydrochloric acid (HCl). Only some ions disintegrate into hydrogen ions (H+) and bicarbonate ions (HCO3-) in a weak acid like carbonic acid (H2CO3), while others are still bound together by ionic bonds.

Define base?

A base is a chemical that, when in solution, emits hydroxyl ions -{OH). We can also define a base as a substance that releases hydroxyl ions (OH-), which mix with any hydrogen ions (H+) in the solution to generate water molecules (OH- + H+ = H2O).

Therefore, a substance that receives or accepts hydrogen ions (H+) that are already present in the solution qualifies as a base.

Because it totally dissociates into sodium ions (Na+) and hydroxyl ions (OH-) when placed in water, sodium hydroxide (NaOH), which is a strong base, is now liberated and dissolves in water.

c

for more information please visite:

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If I use 4 N of force to push an object 3 m, how much work have I done? A)3J B)4N C)12J
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Work done by me for using force of 4 N to push an object 3 m away is 12 J.

<u>Explanation:</u>

Work done on any object is defined as the force required to displace that object from its original position. In other terms, the work done is the measure of force utilized for moving an object. So mathematically it is represented as the product of force with the displacement of the object due to that force.

           Work Done = Force \times Displacement

Then, here it is given that force of about 4 N is used to displace the object to 3m. Thus,

            \text { Work Done }=4 \times 3=12 \mathrm{J}

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Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse
amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given  by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

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