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monitta
3 years ago
10

What is the half-life (in seconds) of a zero-order reaction which has an initial reactant concentration of 0.884 M with a k valu

e of 5.42 × 10–2 M/s?
Chemistry
2 answers:
pantera1 [17]3 years ago
6 0
<span>Answer: 8.15s
</span><span />

<span>Explanation:
</span><span />

<span>1) A first order reaction is that whose rate is proportional to the concenration of the reactant:
</span><span />

<span>r = k [N]
</span><span />

<span>r = - d[N]/dt =
</span><span />

<span>=> -d[N]/dt = k [N]
</span><span />

<span>2) When you integrate you get:
</span><span />

<span>N - No = - kt
</span>
<span></span><span /><span>
3) Half life => N = No / 2, t = t'
</span><span />

<span>=> No - No/ 2 = kt' => No /2 = kt' => t' = (No/2) / k
</span><span />

<span>3) Plug in the data given: No = 0.884M, and k = 5.42x10⁻²M/s
</span>
<span /><span /><span>
t' = (0.884M/2) / (5.42x10⁻²M/s) = 8.15s</span>
lorasvet [3.4K]3 years ago
5 0

Answer:

The half-life (in seconds) of a zero-order reaction is 8.15 seconds.

Explanation:

Initial concentration of the of the reactant = [A_o]=0.884 M

The value of rate constant = k=5.42\times 10^{-2} M/s

The half life for zero order reaction is given as:

t-{\frac{1}{2}}=\frac{[A_o]}{2k}=\frac{0.884 M}{2\times 5.42\times 10^{-2} M/s}=8.15 s

The half-life (in seconds) of a zero-order reaction is 8.15 seconds.

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5 0
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A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0
3 years ago
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