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mixer [17]
3 years ago
11

The decomposition of ethyl amine, C2H5NH2, occurs according to the reaction: C2H5NH2(g)⟶C2H4(g)+NH3(g) At 85∘C, the rate constan

t for the reaction is 2.5 x 10-1 s-1. What is the half-life (in sec) of this reaction?
Chemistry
1 answer:
statuscvo [17]3 years ago
7 0

Answer:

2.772 seconds

Explanation:

Given that;

t1/2 = 0.693/k

Where;

t1/2 = half life of the reaction

k= rate constant

Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant

t1/2 = 0.693/2.5 x 10-1 s-1

t1/2= 2.772 seconds

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How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?
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Answer:

a. Approximately 1.3\; \rm mL.

b. Approximately 7.2\; \rm mL.

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The unit of concentration "\rm M" is equivalent to "\rm mol \cdot L^{-1}", which means "moles per liter."

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\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L.

In a solution of volume V where the concentration of a solute is c, there would be c \cdot V (moles of) formula units of this solute.

Calculate the number of moles of \rm NaCl formula units in each of the two solutions:

Solution in a.:

n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol.

Solution in b.:

n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol.

What volume of that 3.4\; \rm M (same as 3.4 \; \rm mol \cdot L^{-1}) \rm NaCl solution would contain that many

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Convert the unit of that volume to milliliters:

\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL.

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\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL.

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