Hydrogen (H) was first, followed by helium (He).
        
             
        
        
        
Answer:
78 kPa
Explanation:
The total pressure is the sum of the partial pressures:
240 = Pa + Pb + Pc
240 = 107 + 55 + Pc
Pc = 78 kPa
 
        
                    
             
        
        
        
Answer:
V₁  = 10 mL
Explanation:
Given data:
Initial volume of HCl = ?
Initial molarity = 3.0 M
Final molarity = 0.10 M
Final volume = 300.0 mL
Solution:
Formula:
M₁V₁  =  M₂V₂
M₁ = Initial molarity 
V₁  =  Initial volume of HCl 
M₂ =Final molarity
V₂ = Final volume
Now we will put the values.
3.0 M ×V₁  =  0.10 M×300.0 mL
3.0 M ×V₁  = 30 M.mL 
V₁  = 30 M.mL /3.0 M 
V₁  = 10 mL
 
        
             
        
        
        
Answer:
- <u>Yes, it is 14. g of compound X in 100 ml of solution.</u>
Explanation:
The relevant fact here is:
- the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.
That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.
Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.
With that, the solubility can be calculated from the followiing proportion:
- 84. g solute / 600 ml solution = y / 100 ml solution
       ⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.
The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.
<u>The answer is 14. g of solute per 100 ml of solution.</u>
 
        
             
        
        
        
In an average mass, each entry has equal weight. In a weighted average, we multiply each entry by a number representing its relative importance.
Assume that your class consists of 15 girls and 5 boys. Each girl has a mass of 54 kg, and each boy has a mass of 62 kg.
<em>Average mass</em> = (girl + boy)/2 = (54 kg + 62 kg)/2 = <em>58 kg</em>
<em>Weighted average (Method 1)
</em>
Use the <em>numbers of each</em> gender (15 girls + 5 boys)
,
Weighted average = (15×54 kg + 5×62 kg)/20 = (810 kg + 310 kg)/20
= 1120 kg/20 = <em>56 kg</em>.
If you put all the students on one giant balance, their total mass would be
1120 kg and the average mass of a student would be <em>56 kg.
</em>
<em>Weighted average (Method 2)
</em>
Use the <em>relative percentages</em> of each gender (75 % girls and 25 % boys).
Weighted average = 0.75×54 kg + 0.25×62 kg = 40.5 kg + 15.5 kg = <em>56 kg</em>
Each girl contributes 40.5 kg and each boy contributes 15.5 kg to the <em>weighted average</em> mass of a student.