The density can be calculated using the following rule:
density = mass/volume
therefore,
volume = mass/density
we have the mass=22 mg=0.022 grams and density=0.754g/cm^3
substituting in the above equation, we can calculate the volume as follows:
volume = 0.022/0.754 = 0.0291 cm^3

Explanation:
Concentration
;
The reaction see a
change in pressure per second. By the ideal gas law, this quantity would correspond to a change in concentration,
of

where the ideal gas constant
.
By definition, 
Therefore, for this particular reaction

Answer: Graph 3
Explanation:
Activation energy is the level of energy required by a chemical reaction for the reaction to proceed.
A higher Activation energy will translate to a lower reaction rate because less molecules will be able to achieve this required energy needed for them to react. A lower activation energy however will enable a faster reaction time because more molecules will achieve the required energy.
Higher Activation energy = Lower Reaction rate
Lower Activation energy = Higher reaction rate
This is what Graph 3 shows so it is correct.
Answer:
In 0.7 liters (700 ml) there are 122, 85 grams of NaCl
Explanation:
We calculate the weight of 1 mol of NaCl:
Weight 1 mol NaCl= Weight Na + Weight Cl= 23+35,5 g= 58,5g/mol
1 mol NaCl-------58,5 g
3mol NaCl -------x= (3mol NaClx58,5 g)/1 mol= 175, 5 g
3M (175,5 g in this case)---> 3 moles in 1 L of solution
1 L-----175,5 g NaCl
0,7L--x= (0,7Lx175,5 g NaCl)/1L =122,85 g NaCl