gold has a density of 19.3 g/mL. if you have 4.0 grams of gold, what is the volume of the sample?

45 mg = 0.045 gvolume: 0.045 g / (19.3 g/mL) = 0.0023 mLAnswer: 0.0023 mL

Reaction:Au3+ + Fe = Fe3+ + AuV(Au(NO3)3 solution) = 2.8x1023/(6.02x1023)/0.5 = 0.93 LAnswer: 0.93 L

Gold nitrate solution volume =0.93L

30.78 g

A bulk sample is the physical sample of a large volume or tonnage of mineral material taken from a known resource area for investigative purposes.Primary sampling unit refers to Sampling...

You can calculate this by such reaction: NH₄NO₃ --gt; N₂ + frac12;O₂ + 2H₂O ==gt; 2NH₄NO₃ --gt; 2N₂ + O₂ + 4H₂O The law of conservation of mass solves this...

ρ=m/V V=m/ρ V(solid)=27.0/9.0=3 cm3

Solution:gold = AuDensity of a substance is given byDensity = Mass / VolumeMass = Density × VolumeTherefore,(19.3 g Au / 1 mL) × (5 mL) = 96.5 g AuAnswer: 96.5 grams...

p=m/v m=p*v m=737,9*19,3=14241g=14,241kg

V=m/pV=9,2/19,4=0,474=0,5Answer 0,5 см3