Answer
Sounds travel slowest in gasses so it would be air
10. You demonstrated the difference in density of the two objects. It is a physical property.
11. First calculate the density for all of them: density = mass/volume
Density:
A. 5/6 g/ml
B. 10/9 g/ml
C. 15/16 g/ml
D. 20/10 g/ml
If the density of the substance is higher than the density of the substance it is put in, then it will sink. So substances B and D will sink in water, as their densities are higher than 1 g/ml.
12. Ammonia weighs less than water does-- for example, the weight of 8 gallons of ammonia will be equivalent to the weight of 5 gallons of water.
Hope this helped!
Answer:
-3.28 × 10⁴ J
Explanation:
Step 1: Given data
- Pressure exerted (P): 27.0 atm
- Initial volume (Vi): 88.0 L
- Final volume (Vf): 100.0 L
Step 2: Calculate the work (w) done by the gaseous mixture
We will use the following expression.
w = -P × ΔV = -P × (Vf - Vi)
w = -27.0 atm × (100.0 L - 88.0 L)
w = -324 atm.L
Step 3: Convert w to Joule (SI unit)
We will use the conversion factor 1 atm.L = 101.325 J.
-324 atm.L × 101.325 J/1 atm.L = -3.28 × 10⁴ J
There are 2.32 x 10^6 kg sulfuric acid in the rainfall.
Solution:
We can find the volume of the solution by the product of 1.00 in and 1800 miles2:
1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m
1.00 in * 1 m / 39.3701 in = 0.0254 m
Volume = 4.662 x 10^9 m^2 * 0.0254 m
= 1.184 x 10^8 m^3 * 1000 L / 1 m3
= 1.184 x 10^11 Liters
We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70:
[H+] = 10^-pH = 10^-3.7 = 0.000200 M
[H2SO4] = 0.000100 M
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid:
1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4
We can now calculate for the mass of sulfuric acid in the rainfall:
mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
= 2.32 x 10^9 g * 1 kg / 1000 g
= 2.32 x 10^6 kg H2SO4
Answer:
find the number of moles of solute dissolved in solution
,
find the volume of solution in liters,
then divide moles solute by liters solution
Explanation:
