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Rasek [7]
2 years ago
15

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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baherus [9]

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6.673 \times  {10}^{ - 11}

And unit is Nm^2/kg^2

8 0
2 years ago
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What two variables is acceleration dependent on?
Vaselesa [24]
The correct answer to this qustion is velocity and time
7 0
3 years ago
Suppose a thick nimbostratus cloud contains ice crystals and cloud droplets all about the same size. Which precipitation process
Ronch [10]

Answer:The most important process would be the ice crystal process

Explanation:

Ice crystal process also called Bergeron process requires numerous small water drops that are supercooled, which is a common feature in clouds between about 0° and -20°C or below, along with a small number of ice crystals. Therefore because the collision-coalescence process requires that cloud droplets be of varying size so that drops will fill at different speeds, the most important process would be the ice crystal process.

7 0
3 years ago
Eq-45 what type of boat requires navigation lights?
irina [24]

All the boats operating at night requires Navigation light.

Navigation light helps prevent collisions between boats and see if visibility is poor. The types of boat are : Rowboats, Tug boats, Vessels, Sailboat etc.

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8 0
3 years ago
The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso
nalin [4]

Explanation:

It s given that,

Mass of a planet, M=3.7\times 10^{24}\ kg

Radius of a planet, R=9.2\times 10^{6}\ m

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

g=\dfrac{GM}{R^2}

g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}

g=2.91\ m/s^2

(2) The escape velocity is given by :

v=\sqrt{\dfrac{2GM}{R}}

v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}

v = 7324.61 m/s

Hence, this is the required solution.

3 0
3 years ago
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