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olasank [31]
3 years ago
15

In the laboratory you dilute 2.49 ml of a concentrated 6.00 m hydrochloric acid solution to a total volume of 50.0 ml. what is t

he concentration of the dilute solution?
Chemistry
1 answer:
Marizza181 [45]3 years ago
4 0
Equation: M1V1 = M2V2

Where M = concentration & V = volume

Step 1: Write down what is given and what you are trying to find

Given: M1 = 6.00M, V1 = 2.49mL, and V2 = 50.0mL
Find: M2

Step 2: Plug in the values into the equation

M1V1 = M2V2
(6.00M)(2.49mL) = (M2)(50.0mL)

Step 3: Isolate the variable (Divide both sides by 50.0mL so M2 is by itself)

(6.00M)(2.49mL) / (50.0mL) = M2

Answer: M2 = 0.30M 
*Don't forget sig figs & units!
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HELP PLEASE!! ASAP What would the mass be in kg?
fgiga [73]

Answer:

i guess answer is 0.0600

Explanation:

Here,

Density=0.791g/cm^3

Volume=75.89ml

So,

Mass=Density*Volume

=0.791g/cm^3×75.89ml

=60.02g/cm^3.ml

Expressing them in kg

60.02/1000

=0.0600 kg

4 0
3 years ago
12.00 moles of naclo3 will produce how many grams of O2? (Moles to Grams)
Sergeu [11.5K]
2NaClO₃   →  2NaCl  +  3O₂

mole ratio of NaClO₃  to  O₂  is  2  :  3

∴  if moles of NaClO₃  =  12 mol

then moles of O₂  =  \frac{12 mol   *   3}{2}
      
                             =  18 mol


Mass of O₂  =  mol of O₂  ×  molar mass of O₂
 
                    =  18 mol  × 16 g/mol
 
                    =  288 g

So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction.  If you should have used another reaction then just follow the same steps and you'll get your answer.




4 0
3 years ago
How many atoms does 32 grams of sulfur contain?
Lelechka [254]
32 grams of sulfur will contain 6.022 X 1023 sulfur atoms.
4 0
3 years ago
If 8.700 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final tem
leonid [27]
C₆H₆ is benzene which has a molar mass of 78 g/mol. When benzene is burned, the reaction is called combustion. The heat produced in this reaction is called the heat of combustion. For benzene, the heat of combustion is -3271 kJ/mol.

Heat of benzene = (8.7 g)(1 mol/78 g)(-3271 kJ/mol) = -364.84 kJ

By conservation of energy,
Heat of benzene = - Heat of water
where
Heat of Water = mCp(Tf - T₀)
where Cp for water is 4.187 kJ/kg·°C

Thus,

-364.84 kJ = -(5691 g)(1 kg/1000 g)(4.187 kJ/kg·°C)(Tf - 21)
<em>Tf = 36.31°C</em>
8 0
3 years ago
What is the uncertainty of 300.0ft
romanna [79]

Answer:

300.0+\-0.1

Explanation:

3 0
3 years ago
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