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2 years ago

Relative formula mass of CuCO3

Physics

1 answer:

Elodia [21]2 years ago

5 0

AnMolar mass of CuCO3 = 123.5549 g/mol

This compound is also known as Copper(II) Carbonate.

Convert grams CuCO3 to moles or moles CuCO3 to grams

Molecular weight calculation:

63.546 + 12.0107 + 15.9994*3

Percent composition by element

Element Symbol Atomic Mass # of Atoms Mass Percent

Copper Cu 63.546 1 51.431%

Carbon C 12.0107 1 9.721%

Oxygen O 15.9994 3 38.848%

Explanation:

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Select the correct answer.<br> Which hand is negatively charged?

ozzi

Answer:

Explanation:

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3 years ago

9. A wave on Beaver Dam Lake passes by two docks that are 40.0 m apart.

Kobotan [32]

Answers:

a) 10 m

b) time=1.6 s, frquency=0.625 Hz

c) 6.25 m/s

Explanation:

a) If there is a crest at each dock and another three crests between the two docks, and the wavelength $\lambda$ is the distance between to crests; this means we have $4\lambda$ in $40 m$ :

$40 m=4\lambda$

Clearing $\lambda$ :

$\lambda=\frac{40 m}{4}$

$\lambda=10 m$

b) This part can be solved by a Rule of Three:

If 10 waves ---- 16 s

1 wave ----- $T$

Then:

$T=\frac{(1 wave)(16 s)}{10 waves}$

$T=1.6 s$ This is the period of the wave

On the other hand, the frequency $f$ of the wave has an inverse relation with its period $T$ :

$f=\frac{1}{T}$

$f=\frac{1}{1.6 s}$

$f=0.625 Hz$ This is the frequency of the wave

c) The speed $v$ of a wave is given by the following equation:

$v=\frac{\lambda}{T}$

$v=\frac{10 m}{1.6 s}$

Finally:

$v=6.25 m/s$

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3 years ago

A runner begins running at the beginning of 7th period and stops running at the end of the period. She runs at a pace of 10 km/h

MrRissso [65]

Answer:

She run for, t = 0.92 s

Explanation:

Given data,

The velocity of the runner, v = 10 km/h

The distance covered by the runner, d = 9.2 km

The relationship between the velocity, displacement and time is given by the formula,

t = d / v

Substituting the given values in the above equation,

t = 9.2 / 10

= 0.92 s

Hence, she ran for, t = 0.92 s

8 0

3 years ago

Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion

stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 240 K

= initial volume of gas =

= final volume of gas =

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

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2 years ago

Which scale has 100 divisions from when the temperature when water freezes to the temperature when water boils

Westkost [7]

The Celsius scale ( $^{\circ} C$ ).

In the Celsius temperature scale, the temperature at which water freezes is set conventionally at $0^{\circ}C$ , while the temperature at which the water boils is set at $100^{\circ}C$ . The Celsius degree is then defined as the unit corresponding to 1/100 of this time interval, between the temperature of freezing and boiling of the water.

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2 years ago