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anyanavicka [17]
3 years ago
7

40 percent of the fish in a pond are goldfish and the rest are koi. The number of goldfish is g. The farmer then increases the n

umber of Koi by 10 percent. How many Koi are there in the pond, in terms of g, now?​
Chemistry
1 answer:
Stolb23 [73]3 years ago
6 0

Answer:

If 40% are golfish, the Koi are 60% at the start. As it is a calculated ratio, it will apply to any number of original fish

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Answer:

l=3.5*10^{-10}m

Explanation:

From the question we are told that:

1st Energy     E_1=12eV=4(3eV)

2nd Energy  E_2=27eV=9(3eV)

3rd Energy   E_3=48eV=16(3eV)

 

Generally the equation for Energy E for electron in one dimensional box at ground state E_0 is mathematically given by

  E_0=\frac{h}{8ml^2}

  E_0=\frac{h}{8ml^2}

Therefore Length a is mathematically given as

l=\sqrt{\frac{h^2}{8mE_0} }

l=\sqrt{\frac{(6.625*10^{-34})^2}{8(9.1*19^{-31}{(3eV(1.6*10^{-19}))}}

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Answer:

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Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

The balanced molecular equation will be,

Li_2S_{(aq)}+Cu(NO_3)_2_{(aq)}\rightarrow CuS_{(s)}+2LiNO_3_{(aq)}

The complete ionic equation in separated aqueous solution will be,

2Li^+_{(aq)}+S^{2-}_{(aq)}+Cu^{2+}_{(aq)}+2NO_3^{-}_{(aq)}\rightarrow CuS_{(s)}+2Li^+_{(aq)}+2NO_3^{-}_{(aq)}

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