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NNADVOKAT [17]
3 years ago
12

Their are fewer or more elements than compounds?

Physics
2 answers:
zimovet [89]3 years ago
4 0

There are 92 elements that are found in nature.  Roughly 80 of those are capable of combining with each other to make compounds, so there are a lot more than 92 compounds.

Example:  Take carbon and hydrogen.  (please)

Carbon and hydrogen combine to make methane, ethane, propane, pentane, hexane, heptane, octane, butane, and several more 'tanes'.

And if you have some oxygen along with the carbon and hydrogen, you can
make all the alcohols there are, plus vinegar.

EleoNora [17]3 years ago
4 0
There are fewer elements than compounds. There are 118 known elements, but millions of compounds and more being made every day
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bazaltina [42]

Answer:

V=16.65 m^3

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

R=\frac{25}{100}*74kg=18.5kg

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F_b=18.5kg*g

Now the density of the both gases air and helium are different however the volume is the same change offcorss the mass so:

P_h=\frac{m}{V}=0.179 kg/m^3

P_A=1.29 kg/m^3

F_b=F_A-F_H

F_b=m_a*g-m_h*g

m=P/V

18.5kg*g=(1.29kg/m^3-0.179kg/m^3*)V*g

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4 0
3 years ago
Apparent magnitude depends mainly upon _________. our sun has a large apparent magnitude.
Ksenya-84 [330]
Apparent magnitude depends mainly on the brightness of the object as seen from an observer on Earth. This is taken into account without the effects of the atmosphere.
7 0
3 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

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3 years ago
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lana [24]

Answer:

The ball hits the bat

Explanation:

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