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Nadya [2.5K]
3 years ago
8

A beaker contains 20.0 ml of a 50.0 g/L solution. What will the new concentration be if you add 40.0 mL of water?

Chemistry
1 answer:
Ronch [10]3 years ago
8 0

Answer:

moles SO42- = 0.0500 L x 0.20 M=0.010

moles Ba2+ = 0.0500 L x 0.10 M = 0.0050

Ba2+ + SO42- = BaSO4 (s)

moles SO42- in excess = 0.010 - 0.0050=0.0050

total volume = 100 mL = 0.100 L

[SO42-]= 0.0050/0.100= 0.050 M

[Na2SO4] = 2 /2 = 1 M

moles Na2SO4 = 2 M x 0.500 L = 0.500

mass Na2SO4 = 0.500 x 142 g/mol=71.0 g

moles MgBr2 = 46 /184 =0.25

moles Br- = 0.25 x 2 = 0.50

[Br-]= 0.50 / 0.50 L = 1 M

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B. Metal

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Hope it helps plz mark brainlist :)

5 0
3 years ago
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A buffer is prepared by adding 139 mL of 0.39 M NH3 to 169 mL of 0.19 M NH4NO3. What is the pH of the final solution? (Assume th
Paha777 [63]

Answer:

pH = 9.48

Explanation:

We have first to realize that NH₃ is a weak base:

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and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.

Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:

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Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)

pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52

pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48

The solution is basic which agrees  with NH₃ being  a weak base.

5 0
3 years ago
HELP ASAP PLEASE!
trasher [3.6K]

i think the answer is d not sure tho tried to help the best i can if not d than it is c sorry bout that

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