(e) Which sample contains more atoms: 2.0 g of Fe2O3 or 2.0 g of CaSO4? Support your answer with calculations.


Answer 1

CaSO4. It has 6 atoms while Fe2O3 has 5 atoms


Fe2O3 contains 2 atoms of Iron(Fe) and 3 atoms of Oxygen(O) = 2 + 3 = 5 atoms

CaSO4 contains 1 atom of Calcium(Ca), 1 atom of Sulphur(S) and 4 atoms of Oxygen(O) = 1 + 1 + 4 = 6 atoms

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A chemist prepares a solution of magnesium fluoride MgF2 by measuring out 0.00598μmol of magnesium fluoride into a 50.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's magnesium fluoride solution. Round your answer to 2 significant digits.
A weak acid is titrated with 0.1236 M NaOH. From the titration curve you determine that the equivalence point occurs at 12.42 mL of added NaOH. What volume of added NaOH corresponds to the half-equivalence point?

1. How do you determine how many protons are in a neutral element?



The number of protons in an atom is always equal to the atomic number.


For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomassie Blue followed by the addition of 50 ml of 85% v/v phosphoric acid. This entire mixture is then diluted to 1 liter with water. What is the final concentration of phosphoric acid?



4,25% v/v H3PO4


The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:

%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%

In other words, we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition. Which in this case is 1 L (1000 mL).

We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:



Ci, is the initial concentration of the substance.

Vi, the initial volume of the substance

Cf, the final concentration reached after dilution

Vf, the final volume of the solution at which the substance was diluted

In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:

Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v

Note that being up and down in the division, the mL unit is canceled to result in% v / v.

Identify the catalyst-containing systems below as homogeneous or heterogeneous. Catalysts (4 items) (Drag and drop into the appropriate area below) NO(g) for the oxidation of SO2(g) to SO3(g)KI(aq) for the decomposition of H2O2(aq)H2(g) bonded to a metal surfacePd(s) coating that converts noxious gas to less harmful gases


Answer:Homogeneous catalysts:NO(g), KI(aq.)

              Heterogeneous catalyst:H₂(g) bonded to a metal surface,Pd(s) Coating


Homogeneous catalysis is a catalysis reaction in which the reactants as well the catalyst catalyzing the reaction are in the same phase that is, the catalyst and reactants have same phase.

In the examples NO(g) which acts as a catalyst for the oxidation of SO₂(g) to SO₃(g) is in the same gaseous phase  as the reactant SO₂(g). So this is an example of homogeneous catalysis.

In the example KI(aq.) which acts as a catalyst for the decomposition of H₂O₂(aq.) is in the same aqueous phase as reactant H₂O₂(aq.). So this is an example of homogeneous catalysis.  

Hence NO(g) and KI(aq.) are homogeneous catalysts.

Heterogeneous catalysis is a catalysis reaction  in which the reactants and catalyst are in different phase . That is the catalyst and reactant do not have same phase.

In the examples H₂(g) bonded to a metal surface is an example of heterogeneous catalysis as the phase of H₂(g) and metal surface(s) are different hence this is an example of heterogeneous catalysis.

The Pd(s) acting as a catalyst for the conversion of  noxious gas to less harmful gas  is also an example of heterogeneous catalyst as the phase of Pd(s) and gases are different. Pd(s) is solid and the gases are gaseous in nature.

What is the density of iron if it crystallizes in a body-centered cubic unit cell with an edge length of 287 pm



Density = 7.87g/cm^3


Density is the ratio of mass of the given object to the volume of the object, in this question iron is the given object, then we make use of atomic number of iron


Length= 287pm = 287*10^-10cm

Atomic mass of Fe= 56.0u

Z=2(for body centered cubic unit cell)

Avogadro number (N 0)=6.022× 10^23

Density= ZM/a^3 × N


Z= body centered cubic unit cell

Then substitute

N= Avogadro's number


Density = (2× 56)/(287*10^-10cm)^3 × (6.022 × 10^23)

Density = 7.87g/cm^3

Final answer:

The density of iron in a body-centered cubic unit cell can be calculated using the mass and volume of the unit cell.


The density of iron can be calculated using the formula: density = mass/volume. To determine the mass of the unit cell, we need to know the molar mass of iron and the number of atoms in the unit cell. The molar mass of iron is 55.845 g/mol, and there are two iron atoms in the body-centered cubic unit cell of iron. The volume of the unit cell can be calculated using the formula: volume = (edge length)^3.

Putting these values into the formula, we get:

density = (2 * 55.845 g/mol) / ((287 pm)^3)

Converting the edge length to meters (1 pm = 1e-12 m) and calculating, we find that the density of iron is approximately 7.86 g/cm³.

Learn more about density of iron here:



The higher the pH, the less acidic the solution


If the pH is higher the concentration of hydrogen ions becomes less and the solution becomes less acidic.

As the pH becomes lower, the concentration of hydrogen ions becomes greater, and the solution becomes more acidic.



yh thats true lol, ty for that very interesting fact

Select the correct answer.In which research method does a researcher change the value of one variable to determine how the change affects another variable?
O A experiment
OB. observational research
case study



experiment is the answer