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Korvikt [17]
3 years ago
12

How many moles are in 2.5L of 1.75 M Na2CO3

Chemistry
1 answer:
mixas84 [53]3 years ago
6 0

Answer: There are 4.375 moles in 2.5 L of 1.75 M Na_2CO_3

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}      

Molarity of solution = 1.75 M

Volume of solution = 2.5 L  

Putting values in equation , we get:

1.75M=\frac{\text{Moles of} Na_2CO_3}{2.5L}\\\\\text{Moles of }Na_2CO_3=1.75mol/L\times 2.5L=4.375mol

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Free-energy change, ΔG∘, is related to cell potential, E∘, by the equationΔG∘=−nFE∘where n is the number of moles of electrons t
mart [117]

Answer:

a)\Delta G=372490 J

b)\Delta G=-568614 J

Explanation:

a) The reaction:

Mg(s) +Fe^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Fe(s)

The free-energy expression:

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E=E_{red}-E_{ox]

The element wich is reduced is the Fe and the one that oxidates is the Mg:

-0.44V=E_{red}-(-2.37V)=1.93V

The electrons transfered (n) in this reaction are 2, so:

\Delta G=-2mol*96500 C/mol * 1.93 V

\Delta G=-372490 J

b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

\Delta G=\Delta H - T* \Delta S

with T in Kelvin

\Delta G=-675 kJ*\frac{1000J}{kJ} - 298K*-357 J/K

\Delta G=-568614 J

7 0
3 years ago
Consider the following reaction.
Ivan

Answer:

162 g Fe₂O₃

Explanation:

To find the mass of Fe₂O₃, you need to (1) convert grams C to moles C (via molar mass from periodic table), then (2) convert moles C to moles Fe₂O₃ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe₂O₃ to grams (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the given value.

Molar Mass (C): 12.011 g/mol

2 Fe₂O₃(s) + 3 C(s) ---> 4 Fe(s) + 3 CO₂(g)

Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)

Molar Mass (Fe₂O₃): 159.684 g/mol

18.3 g C           1 mole            2 moles Fe₂O₃          159.684 g
--------------  x  ----------------  x  -------------------------  x  -----------------  = 162 g Fe₂O₃
                        12.011 g              3 moles C                 1 mole

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Answer:

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A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added
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Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄

On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.

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3 years ago
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