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tester [92]
3 years ago
10

Mr. Rutherford's chemistry class was collecting data in a neutralization study. Each group had 24 test tubes to check each day f

or one week. Students had to check to see whether the solution in each test tube was acidic, basic, or neutral. What would be the MOST efficient method for collecting this data and recording it for later analysis? A) pH paper and a notebook B) pH probe linked to a graphing calculator C) photometer linked to a graphing calculator D) phenolphthalein solution and a computer spreadsheet
Chemistry
1 answer:
ycow [4]3 years ago
6 0
I’m pretty sure it is A at least that’s what we did at our school to test this
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Suppose 180 ml of 3.52x10^-4 M NaOH is mixed with 220 mL of 2.47x10^-4 M MgCl2.
velikii [3]
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:

MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-

The initial concentrations are as follows:

[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
6 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
Water is being pumped from the bottom of a well 150 feet deep at a rate of 200 gal/hour into a vented storage tank 30 feet above
KengaRu [80]

Explanation:

As the given data is as follows.

        Height, H = 150 feet

 Heat gain = 30,000 BTU/hr,  and  Heat loss = 25000 BTU/hr

  m = mass of water heated = 700 gallons = 5810 lbs

C_{p} is the heat capacity of water = 1 BTU/lb ^{o}F (given)

      \Delta T = temperature difference = 120^{o}F - 35^{o}F

Heat energy required to heat 700 gal can be calculated as follows:

    Heat Required = 5810 lbs \times 1 BTU/lb^{o}F \times (120^{o}F - 35^{o}F)

Thus, water rises till 120^{o}F.

3 0
3 years ago
Question 6 (1 point)
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This statement would be best characterized by the law of conservation of momentum—choice C.
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Which enzyme is responsible for facilitating the hydrogen bonding
jeka57 [31]

Answer:

its the polymerase

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