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3 years ago

7.26 of a hydrate, Cu(NO3)2.xH2O, formed 2.4 g copper(II) oxide.

Chemistry

1 answer:

goldfiish [28.3K]3 years ago

6 0

Number of moles= mass/ molar mass

Or n=m/MM

n = number of moles

m = mass

MM = molar mass

1) n CuO = 2.4g / 79.54g/mol = 0.03 mol CuO

2) n Cu(NO3)2.xH2O = 7.26 g / 205.6 = 0.035 moles of Cu(NO3)2.xH2O

3) 205.6 g

Cu = 63.5 g

N = 14g

O =16g

H= 1 g

63.5+ (14+(16*3))*2+1*2+16 =205.6 g

4) yes is 188g

5) I don’t know, I assume was 1

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A pump with an 80% efficiency drives water up between two reservoirs through a piping system of total length L = 15 and circular

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Length of the pipe, L = 15 m

Cross-sectional diameter, d = 7 cm

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Therefore;

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Which gives;

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