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3 years ago

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Which equation is used to determine the density of a substance

1 answer:

9966 [12]3 years ago

8 0

Density = (mass)/(volume)

IF 'D' means density, 'M' means mass, and 'V' means volume, then choice-A is the equation to use.

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Estimate the number of atoms in 1 cm^3 of a solid

dexar [7]

Avogadro's number: 6.02 x 10^23 atoms is present in 1mol of a solid (i.e. 22, 400 cm3)

Hence, in 1 cm3, 6.02 x 10^23 /22400 atoms is present = 2 x 10 ^ 19 atoms.

7 0

3 years ago

1. What is technology?<br>techndegy

Naya [18.7K]

Answer:

the application of scientific knowledge for practical purposes, especially in industry. Another answer:the sum of techniques,skills,methods and processes used in the production of goods or services or in the accomplishment of objectives, such as scientific investigation

6 0

4 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

4 years ago

True Or False: people in the U.S. consume more liters of water than those in developing nations.

Andru [333]

Truee hope this helps

7 0

3 years ago

If two opposite charges get farther apart, does their attraction for each other increase, decrease, or remain the same?

wlad13 [49]

The force between them <em>decreases</em><em>,</em> as the square of the distance.

For example ...

-- If you move them apart to double the original distance, the force becomes (1/2²) = 1/4 of the original force.

-- If you move them apart to 3 times the original distance, the force becomes (1/3²) = 1/9 of the original force.

-- If you move them apart to 5 times the original distance, the force becomes (1/5²) = 1/25 of the original force.

(Gravity works exactly the same way.)

7 0

3 years ago