Taking into account the definition of density, the density of the sample is 0.5 
.
It is necessary yo know that density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.
In other words, density allows you to measure the amount of mass in a certain volume of a substance.
Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:
In this case, you know:
Then, replacing in the definition of density:
Solving:
<u><em>density= 0.5 </em></u><u><em /></u>
Finally, the density of the sample is 0.5 .
Learn more about density:
Answer:
hope it helps...
Explanation:
Along the third type of plate boundary, two plates move laterally and pass each other along giant fractures in Earth's crust. Transform faults are so named because they are linked to other types of plate boundaries. The majority of transform faults link the offset segments of oceanic ridges.
I know the answer to your question but if you want me to answer this question first, I need your help with my question "How do I solve number 12? Quick please I need help ASAP!!!"
Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
