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Viktor [21]
4 years ago
13

A chemist prepares a solution of barium acetate BaCH3CO22 by measuring out 81.4μmol of barium acetate into a 150.mL volumetric f

lask and filling the flask to the mark with water. Calculate the concentration in /mmolL of the chemist's barium acetate solution. Round your answer to 3 significant digits.
Chemistry
1 answer:
grigory [225]4 years ago
5 0

Answer:

The answer to your question is Concentration = 0.54 mmol/L

Explanation:

Data

Barium acetate = BaCH₃CO₂

mass = 81.4 μmol

volume = 150 ml

concentration = mmol / L

Process

1.- Convert the μmol to mmol

                        1000 μmol ------------------ 1 mmol

                           81.4 μmol ----------------- x

                           x = (81.4 x 1) / 1000

                           x = 0.0814 mmol of BaCH₃CO₂

2.- Convert the volume to liters

                           1000 ml -------------------- 1 l

                             150 ml --------------------- x

                              x = (150 x 1) / 1000

                              x = 0.15 l

3.- Calculate the concentration

Concentration = 0.0814 / 0.15

                        = 0.54 mmol/L

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Mila [183]

Answer:

The correct answer is nuclear binding energy.

Explanation:

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5 0
4 years ago
What type of bond is SF6
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Hope this helps:)

3 0
3 years ago
Which of the following is included in nuclide symbols, but is not strictly necessary for the identification of the nuclide?
kodGreya [7K]

Answer:

c. isotope number

Explanation:

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The correct option is option C. This is the symbol that is not necessary for the identification of a nuclide.

5 0
3 years ago
Kp/Kc for reaction for the equilibrium, A(g) ⇌ C(g)+B(g), is _______.
Sophie [7]

Kp/Kc = RT

<h3>Further explanation</h3>

Given

Reaction

A(g) ⇌ C(g)+B(g)

Required

Kp/Kc

Solution

For reaction :

pA + qB ⇒ mC + nD  

\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}

While the equilibrium constant Kp is based on the partial pressure  

\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}

The value of Kp and Kc can be linked to the formula '  

\large {\boxed {\bold {Kp ~ = ~ Kc. (R.T) ^ {\Delta n}}}}

R = gas constant = 0.0821 L.atm / mol.K  

Δn=moles products - moles reactants or

number of product coefficients-number of reactant coefficients  

For reaction :

A(g) ⇌ C(g)+B(g)

number of product coefficients = 1+1=2

number of reactant coefficients   = 1

Δn= 2 - 1 =1

So Kp/Kc = RT

5 0
3 years ago
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ivann1987 [24]
Hello!

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5 0
4 years ago
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