A cube of wood having an edge dimension of 18.0 cm and a density of 651 kg/m3 floats on water.(a) What is the distance from the horizontaltop surface of the cube to the water level?(b) How much lead weight has to be placed on top of the cube sothat its top is just level with the water?

(a) The distance will be "6.282 cm".

(b) Mass will be "2.03 kg".

(a)

From Archimedes' principle, we get

→

So,

The distance from horizontal top to the water level will be:

=

=

(b)

After placing the lead block of weight , we have

→

Mass,

m = 2.03 kg  

Thus the above answer is right.

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A. 6.282

B. 2.03kg

Explanation:

A.

We solve using archimedes principle

L³pwood = L²dwater

We make d subject of the formula

d = Lpwood/pester

= 18x651/1000

= 18x0.651

= 11.718cm

Distance from horizontal top to water level

= 18-11.718

= 6.282cm

B.

When we place lead block

WL + L³pwoodg = L³pwaterg

WL = L³g(Pwater-Pwood)

= 0.18³x9.8(1000-651)

= 19.94N

19.94/9.8

= 2.03kg

The mass m is therefore 2.03kg

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz