M=D*V
D=620 g/cm³
V=75 cm³
m= 620 g/cm³ * 75 cm³=46500 g
m=46500g
Answer:
C 8.09 SO2 gas
Explanation:
As we have the volume (3dm³ = 3L), temperature (25°C + 273 = 298K), and pressure (1atm), we can solve to moles of gas using:
PV = nRT
PV / RT = n
1atm*3L / 0.082atmL/molK*298K =¨
0.123 moles of gas you have.
Now, to convert these moles to mass we use molar mass (32g/mol for O2, 28g/mol for N2, 64g/mol for SO2, and 44g/mol for CO2).
Mass of 0.123 moles of these gases is:
O2 = 0.123 moles * 32g/mol = 3.94g of O2. A is wrong
N2 = 0.123 moles * 28g/mol = 3.4g of N2. B is wrong
SO2 = 0.123 moles * 64.1g/mol = 7.9g of SO2≈ 8.09g of SO2, C is possible
CO2 = 0.123 moles * 44g/mol = 5.4g of CO2. D is wrong
Right answer is:
<h3>C 8.09 SO2 gas
</h3>
Net Primary Productivity ... the amount of biomass present in an ecosystem at a particular time .... Explain why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.
Answer : Both solutions contain
molecules.
Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain
molecules.
Avogadro's Number is
=
which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.
Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.
Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.
Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.
We can calculate the number of molecules for each;
Number of molecules =
;
∴ Number of molecules =
which will be = 
Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.