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Arte-miy333 [17]
3 years ago
14

Which of the following is a property of an acid

Chemistry
2 answers:
Pavel [41]3 years ago
5 0

Answer:

fating on acid and bases will give u the property of paper blue litmus and they tase sour and they typically contain an =oh group with slippery stuff and a propety of acids

Explanation:

liberstina [14]3 years ago
3 0

Answer:

Explanation: Acids give off H+ (Hydrogen) ions in water; bases give off OH- (Hydroxide) ions in water. Acids generally taste sour due to the sour H+ ion; bases taste bitter due to the OH- ion; but they may have other tastes depending on the other part of the molecule. ... Acids have a pH less than 7 ; Bases have a pH greater than 7.

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D

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Calculate the formula for the following hydrate composed of 76.9% CaSO3 and 23.1%H2O
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Assume there is 100g of the substance at first

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4. An 82 kg hiker climbs Mt. Humphrey near Flagstaff. During a two hour period, the hiker's
bazaltina [42]

Answer:

\Delta E = 434253.96\,J, \Delta P = 60.313\,W

Explanation:

The change in energy is given by the change in gravitational potential energy:

\Delta E = m\cdot g \cdot \Delta h

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The average rate of change in terms of time is approximately this:

\Delta P = \frac{\Delta E}{\Delta t}

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\Delta P = 60.313\,W

8 0
3 years ago
determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

3 0
3 years ago
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horsena [70]

Answer: Option (3) is the correct answer.

Explanation:

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In aerobic respiration, food releases energy to produce ATP which is necessary for cell activity. There is complete breakdown of glucose in aerobic respiration that is why more energy is released. Therefore, aerobic organisms become active.

Thus, we can conclude that characteristics very active, efficient use of energy describes aerobic organisms.

6 0
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