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sattari [20]
3 years ago
15

If you mixed three Dixie cups’ contents together containing 0.05L of 1.0M lemonade, 0.05L of 2.5M lemonade, and 0.05L of 0.5M le

monade, what would be the molarity of the resulting solution?
Chemistry
1 answer:
viva [34]3 years ago
3 0

Molarity of the resulting solution will be 1.33 M.

<u>Explanation:</u>

First we have to find the number of moles for each of the solution using the formula, moles = molarity × volume

For cup 1  = 1  M ×0.05 L = 0.05 moles

For cup 2 = 2.5 M × 0.05 L= 0.125 moles

For cup 3 = 0.5 M × 0.05 L = 0.025 moles

Total moles = 0.05 + 0.125 + 0.025  = 0.2 moles

We have to find the total volume as, 0.05 + 0.05 + 0.05 = 0.15 L

Now we have to find the molarity as, moles / volume = 0.2 moles/ 0.15 L = 1.33 M

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The final temperature : 44.5 °C

<h3>Further explanation</h3>

Heat can be formulated :

Q = m . c .Δt

Q = heat, J

m = mass, g

c = specific heat, J/gC

Δt= temperature

m= 625 g

Q=7.96 x 10⁴ J

c = 4.18 j/gC

t₁=75 C

\tt Q=m.c.\Delta t\\\\7.96\times 10^4=625\times 4.18\times (75-t_2)\\\\75-t_2=30.5\\\\t_2=44.5^oC

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Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2
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Answer:

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Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

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This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

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A balloon contains helium gas expands from 230ml to 860 ml as more helium is added. What was the initial quantity of helium pres
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Answer:

n₁ = 1.0× 10⁻⁴ mol

Explanation:

Given data:

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Solution:

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Mathematical relationship:

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n₁ = 230 mL× 3.8 × 10⁻⁴ mol/ 860 mL

n₁ = 874 × 10⁻⁴ mol. mL / 860 mL

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