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Kruka [31]
3 years ago
6

An anhydrous sample of CaCO3 (100.089 g/mol) weighing 0.9849 g was quantitatively transferred into a 250.0 mL volumetric flask h

alf-full of Nanopure H2O. One milliliter of 12M HCl was added & the solution was swirled. Then the flask was filled to volume with Nanopure H2O. What is the molarity (M) of the solution?
Chemistry
1 answer:
Katyanochek1 [597]3 years ago
4 0

The molarity of CaCO₃ in the final solution is equal to 0.015 M.

Explanation:

When we add HCl over CaCO₃ we have the following chemical reaction:

CaCO₃ + 2 HCl → CaCl₂ + H₂CO₃

H₂CO₃ (is not stable) → CO₂ + H₂O

number of moles = mass / molecular weight

number of moles of CaCO₃ = 0.9849 / 100.089 = 0.0098 moles

molar concentration = number of moles / volume (L)

number of moles = molar concentration × volume (L)

number of moles of HCl = 12 × 0.001 = 0.012 moles

From the reaction we see that 1 mole of CaCO₃ will react with 2 moles of HCl so the limiting reactant will be HCl. Knowing this we formulate the following reasoning:

if         1 mole of CaCO₃ will react with 2 moles of HCl

then   X moles of CaCO₃ will react with 0.012 moles of HCl

X = (0.012 × 1) / 2 = 0.006 moles of CaCO₃

The number of moles of CaCO₃ which remain unreacted are equal to 0.0098 (initial quantity) - 0.006 (reacted with HCl) = 0.0038 moles.

Now the molarity of CaCO₃ in the final solution:

molar concentration = number of moles / volume (L)

molar concentration of CaCO₃ = 0.0038 / 0.250 = 0.015 M

Learn more about:

molar concentration

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Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).
muminat

Hello!

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

We have the following data:

m (mass) = ? 

n (number of moles) = 5.20 moles

MM (Molar mass of C6H12) ≈ 84.2 g/mol

Now, let's find the mass, knowing that:

n = \dfrac{m}{MM}

5.20\:\:\diagup\!\!\!\!\!\!\!mol = \dfrac{m}{84.2\:g/\diagup\!\!\!\!\!\!\!mol}

m = 5.20*84.2

\boxed{\boxed{m = 437.84\:g}}\end{array}}\qquad\checkmark

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I Hope this helps, greetings ... Dexteright02! =)

4 0
3 years ago
Read 2 more answers
If nitrogen-13 has a half life of 2.5 years, how much remains from a 100g sample after 7.5 years
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Answer:

12.50g

Explanation:

T½ = 2.5years

No = 100g

N = ?

Time (T) = 7.5 years

To solve this question, we'll have to find the disintegration constant λ first

T½ = In2 / λ

T½ = 0.693 / λ

λ = 0.693 / 2.5

λ = 0.2772

In(N/No) = -λt

N = No* e^-λt

N = 100 * e^-(0.2772*7.5)

N = 100*e^-2.079

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Let's convert the volume of solution in L

10 mL . 1L/1000 mL = 0.01 L

4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L

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Answer:

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