Answer:
B. The elements that have similar densities and atomic sizes
Explanation:
1) Dawn dish soap has a density of 1.06 g/mL. If the mass of a sample of the liquid is 1.00 g what is the volume?
Answer:
v = 0.94 mL
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Given data:
Density of soap = 1.06 g/mL.
Mass = 1 g
Volume = ?
Solution:
d = m/v
v = m/d
v = 1 g/1.06 g/mL
v = 0.94 mL
2) Maple syrup has a density of 1.37 g.mL. What is the mass of 1.0 L of the maple syrup?
Answer:
m = 1370 g
Given data:
Density of soap = 1.37 g/mL.
Mass = ?
Volume = 1.0 L ( 1000 mL)
Formula:
D=m/v
D= density
m=mass
V=volume
Solution:
d = m/v
m = d × v
m = 1.37 g/mL × 1000 mL
m = 1370 g
3) The density of gasoline is 0.754 g/mL. A drop of gasoline has a mass of 22 g what is the volume?
Answer:
v = 29.2 mL
Given data:
Density of soap = 0.754 g/mL.
Mass = 22 g
Volume = ?
Formula:
D=m/v
D= density
m=mass
V=volume
Solution:
d = m/v
v = m/d
v = 22 g/0.754 g/mL
v = 29.2 mL
Answer:
it burns things which would be burned easily by lightning, and then people put the fire out immediately. If the dried plants that they are lighting on fire are hit by lightning, it can lead to a massive fire without anyone realizing.
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp =
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104![[P(NO_{2} )]^{2}](https://tex.z-dn.net/?f=%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D)
+ 
- 200 = 0
Resolving the second degree equation:
= 
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are = 98.7 MPa and P(N₂O₄) = 101.3 MPa
