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:
Answer:
50 W
Explanation:
Case 1
Power = V * I
100 = 220 * I
I =
A
Case 2
P = V * I
P = 110 * 
P = 50 W
I think the answer is 50 W
Hope it helps
Once energy from the Sun gets to Earth, several things can happen to it:
Energy can be scattered or absorbed by aerosols in the atmosphere. Aerosols are dust, soot, sulfates and nitric oxides. When aerosols absorb energy, the atmosphere becomes warmer. When aerosols scatter energy, the atmosphere is cooled.
Short wavelengths are absorbed by ozone in the stratosphere.
Clouds may act to either reflect energy out to space or absorb energy, trapping it in the atmosphere.
The land and water at Earth's surface may act to either reflect energy or absorb it. Light colored surfaces are more likely to reflect sunlight, while dark surfaces typically absorb the energy, warming the planet.
Albedo is the percentage of the Sun's energy that is reflected back by a surface. Light colored surfaces like ice have a high albedo, while dark colored surfaces tend to have a lower albedo. The buildings and pavement in cities have such a low albedo that cities have been called "heat islands" because they absorb so much energy that they warm up.
The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
The given parameters;
- <em>initial temperature of metals, = </em>
<em /> - <em>initial temperature of water, = </em>
<em> </em> - <em>specific heat capacity of copper, </em>
<em> = 0.385 J/g.K</em> - <em>specific heat capacity of aluminum, </em>
= 0.9 J/g.K - <em>both metals have equal mass = m</em>
The quantity of heat transferred by each metal is calculated as follows;
Q = mcΔt
<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
Learn more here:brainly.com/question/15345295
To solve this problem we will start from the given concept in which the number of turns is equivalent to the length of the thread per circumference of spool. That is:

Where,
l = length of the thread
= circumference of spool
For \phi we have that,

For l we have that
l = 62.8m
Finally the number of Turns would be,



Therefore the number of turns of thread on the spool are 1000turns.