Answer:
Percent error = 12.5%
Explanation:
In a measurement you can find percent error following the formula:
Percent error = |Measured value - Accepted Value| / Acepted value * 100
Based on the data of the problem, accepted value is 22.4L and the measured Value (Value of Sara) was 19.6L.
Replacing:
Percent error = |Measured value - Accepted Value| / Acepted value * 100
Percent error = |19.6L - 22.4L| / 22.4L * 100
Percent error = |-2.8L| / 22.4L * 100
Percent error = 2.8L / 22.4L * 100
Percent error = 12.5%
 
        
             
        
        
        
Answer:
12.10 mol / 1 L 
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity ( unit = mol / L or M )
V = volume of solution in liter ( unit = L ),  
n = moles of solute ( unit = mol ),  
Moles is denoted by given mass divided by the molecular mass ,  
Hence , 
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .  
From the question ,
The data given is as follows -
w = 439 g
As , we known for HCl , 
m = 36.46 g/mol
V = 1 L 
From the above data ,
Moles are given as -
n = w / m
n = 439 / 36.26 = 12.10 mol ,
Now , the molarity is given as ,
M = n / V
M = 12.10 mol / 1 L 
M = 12.10 mol /L
 
        
             
        
        
        
Take the atomic mass of silicon and put it over one. Then set that equal to x over 4.8 x 1026. X will equal to the weight of silicon in grams.
        
             
        
        
        
Answer:
13.5 moles of AgNO₃
Explanation:
To determine the reaction:
Reactants: AgNO₃ and Cu
Products: Cu(NO₃)₂ and Ag
2 moles of AgNO₃ react to 1 mol of Cu, in order to produce 1 mol of Cu(NO₃)₂ and 2 moles of solid silver.
2AgNO₃ + Cu → Cu(NO₃)₂ + 2Ag
Our production was 6.75 moles of Cu(NO₃)₂
Let's make the rule of three:
1 mol of Cu(NO₃)₂ is produced by 2 moles of AgNO₃
Then, our 6.75 moles were definetely produced by (6.75 . 2) /1 = 13.5 moles.
If the copper was in excess, then the silver nitrate is the limiting reactant:
2 mol of AgNO₃ can produce 1 mol of Cu(NO₃)₂
Then, 13.75 moles of silver nitrate must produce (13.5 . 1) /2 = 6.75 moles of Cu(NO₃)₂