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VARVARA [1.3K]
4 years ago
12

(a) With a of specific gravity 1.42 were 2.8 min and 5.2 minutes, respectively. The experiment was carried out at 30°C. Calcula

te the viscosity of the second liquid. Viscosity of water at 30°C = 0.80 cP; density of water at 30°C 0.9956 g/mL (b) A sphere of diameter 1.8 mm and density 2.75 g/cm2 falls at a constant velocity through large volume of the second liquid in (a). Calculate the time required for the ball to fall a distance of 90 cm through the second liquid given Ostwald viscometer, it was found that the times of flow of water and a second liquid
Chemistry
1 answer:
mr Goodwill [35]4 years ago
3 0

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The game of kickball begins with the ball being rolled toward a player. The player then kicks the ball across the field. What is
hammer [34]

Answer:

C. The Speed of the ball depends on the force used to kick it.

Explanation:

6 0
3 years ago
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Can anyone check my work and see if it is correct? If not, may someone help me?
shusha [124]
It looks all correct to me, great job!
6 0
3 years ago
The volume of carbon gas at 5.0 ATM was measured to be 363 ml. What will the pressure be if the volume is change to .00020 ml?
aleksley [76]

Answer:

The answer to your question is P2 = 9075000 atm

Explanation:

Data

Pressure 1 = P1 = 5 atm

Volume 1 = V1 = 363 ml

Pressure 2 = P2 = ?

Volume 2 = 0.0002 ml

Process

To solve this problem use Boyle's law

                 P1V1 = P2V2

-Solve for P2

                 P2 = P1V1/V2

-Substitution

                  P2 = (5 x 363) / 0.0002

-Simplification

                  P2 = 1815 / 0.0002

-Result

                 P2 = 9075000 atm

5 0
4 years ago
The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?
Aloiza [94]

Answer:

0.0468 g.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>

8 0
3 years ago
Perform the following mathematical operation, and report the answer to the correct number of significant figures. 0.34 x 0.568=?
slavikrds [6]

The correct answer to the problem is 0.193 which is three significant figures.

<h3>What are significant figures?</h3>

The term significant figures has to do with the figures that have a mathematical meaning. We know that the result has to correspond to the highest number of significant figures.

Hence, If we multiply 0.34 x 0.568 the result ought to be recorded as 0.193 which is three significant figures.

Learn more about significant figures:brainly.com/question/14804345

#SPJ1

6 0
2 years ago
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