This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.
<h3>Dimensional analysis</h3>
In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.
Thus, since this problem asks for try ounces in an average Nevada ore, which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:
Where the short tons are cancelled out as well as the grams, in order to obtain:
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The concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 M dextrose solution to 25 ml using a 25 ml volumetric flask is 0.56M.
Concentration is defined as the number of moles of a solute present in the specific volume of a solution.
According to the dilution law, the degree of ionization increases on a dilution and it is inversely proportional to the square root of concentration. The degree of dissociation of an acid is directly proportional to the square root of a volume.
M₁V₁=M₂V₂
Where, M₁=1.0M, V₁=14ml, M₂=?, V₂=25ml
Rearrange the formula for M₂
M₂=(M₁V₁/V₂)
Plug all the values in the formula
M₂=(1.0M×14 ml/25 ml)
M₂=14 M/25
M₂=0.56 M
Therefore, the concentration of a dextrose solution after the dilution is 0.56M.
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4.88x10^20 H2O2 molecules
Answer:
the cell can be the described as the basic unit of life
The mixture contains:
CaCO3 + (NH4)2CO3 in which the amount of carbonate CO3 = 60.7% by mass
Let, the total mass = 100 grams
Mass of CaCO3 = x grams
Mass of (NH4)2CO3 = y grams
Thus, x + y = 100 ------------(1)
Mass of CO3 = 60.7% = 60.7 g
Molar mass of CO3 = 60 g/mol
Total # moles of CO3 = 60.7 g/60 g.mol-1 = 1.012 moles
The total moles of CO3 comes from CaCO3 and (NH4)2CO3. Therefore,
moles CaCO3 + moles (NH4)2CO3 = 1.012
mass CaCO3/molar mass CaCO3 + mass (NH4)2 CO3/molar mass = 1.012
x/100 + y/96 = 1.012---------(2)
based on equation 1 we can write: y = 100-x
x/100 + (100-x)/96 = 1.012
x = 71.2 g
Mass of CaCO3 = 71.2 g
