Aspring stretches by 0.0247 m when a 4.78-kg object is suspended from its end. how much mass should be attached to this spring so that its frequency of vibration is f

This question is incomplete.So complete question is given as

I have assumed value of f=6 Hz

A spring stretches by 0.0247 m when a 4.78-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f=6 Hz (assumed value)

mass=4.194 kg

Explanation:

x (Spring stretches)=0.0247 m

m(mass)=4.78 kg

f=6 Hz (assumed)

mass required attached this spring=?

Solution

  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