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larisa86 [58]
3 years ago
7

HELPPPP. SHOW WORK

Chemistry
1 answer:
nekit [7.7K]3 years ago
7 0
I think The answer to two is it would sink
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Which of the following will allow measurement of a liquids volume with the greatest precision ?​
Evgesh-ka [11]
I personally think that it’s C.
(i’m sorry if it’s wrong)
6 0
3 years ago
Read 2 more answers
What is the overall enthalpy of reaction for the equation shown below?
Rudiy27

Answer:

ΔH₁₂ = -867.2 Kj

Explanation:

Find enthalpy for 3H₂ + O₃ => 3H₂O given ...

2H₂ + O₂ => 2H₂O      ΔH₁ = -483.6 Kj

        3O₂ => 2O₃        ΔH₂ = + 284.6 Kj

_____________________________

3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O       (multiply by 3 to cancel O₂)

6H₂ + 3O₂ => 6H₂O        ΔH₁ = 3(-483.6 Kj) = -1450.6Kj

          2O₃ => 3O₂           ΔH₂ = -284.6Kj              (reverse rxn to cancel O₂)

_______________________________

6H₂ + 2O₃ => 6H₂O         ΔH₁₂ = -1735.2 Kj       (Net Reaction - not reduced)

________________________________

divide by 2 => target equation (Net Reaction - reduced)

3H₂ + O₃ => 3H₂O            ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj    

4 0
3 years ago
A scuba diver with a local Sheriff’s office has been assigned the task of retrieving a sunken vessel in the Ocean. The vessel is
n200080 [17]
First, we convert the depth of the water into meters. This is:
60 feet = 18.3 meters

Now, we compute the additional pressure exerted due to the water, which is given by:
Pressure = density * gravitational field strength * height
P = 1000 * 9.81 * 18.3
P = 179.5 kPa

The atmosphere pressure is 101.325 kPa
The pressure of the gas bubbles 60 feet under water will be:
179.5 + 101.325 = 280.825 kPa

The pressure at the surface of the water will be equal to the atmospheric pressure, 101.325 kPa.

Because of this decrease in external pressure as gas bubbles rise, they are seen to expand.
4 0
3 years ago
Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0
3 years ago
a solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. what is the freezing point depression of
Lesechka [4]
The freezing point depression of the solution or pure substance that is added with the solvent is calculated through the equation,
    
  ΔTf = Kfm

where ΔT is the freezing point depression, Kf is the constant for water given to be -1.86°C/m and m is the molality of the solution. 

Molality is calculated through the equation,
    
    m = number of moles solute/ kg of solvent

Calculation of molality is shown below.
    
   m = (21.5 g C6H12O6)(1 mol/180 g) / (0.255 kg)
     m = 0.468 molal

The freezing point depression is then,
    
   ΔTf = (-1.86°C/m)(0.468 m) = -0.87°C

<em>Answer: -0.87°C</em>
3 0
3 years ago
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