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sladkih [1.3K]
3 years ago
14

A first-order reaction has a rate constant of 0.241/min. if the initial concentration of a is 0.859 m, what is the concentration

of a after 10.0 minutes?
Physics
2 answers:
nordsb [41]3 years ago
5 0

Answer: 0.077 M

Explanation:

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

where,

k = rate constant = 0.241minute^{-1}

t = time taken for decay process = 10 minutes

a = initial amount of the reactant= 0.859 M

a - x = amount left after decay process =?

Putting values in above equation, we get:

0.241 minutes^{-1}=\frac{2.303}{10.0}\log\frac{0.859}{a-x}

(a-x)=0.077M

Thus the concentration of a after 10.0 minutes is 0.077 M.


Juli2301 [7.4K]3 years ago
3 0
Initial concentration is 0.859 m.
A rate: 0.241 m/min.
In 10 minutes: 10 min * 0.241 m/min = 2.41 m
0.859 m + 2.41 = 3.269 m
Answer: After 10 minutes the concentration is 3.269 m. 
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Explanation:

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a - ?

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Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain
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Answer:

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Explanation:

Given:

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enthalpy of vaporization of water, h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}

specific heat of water, c=4180\ J.kg^{-1}.K^{-1}

Amount of heat required to raise the temperature of given water mass to 100°C:

Q_s=m.c.\Delta T

Q_s=1\times 4180\times (100-18)

Q_s=342760\ J

Now the amount of heat required to vaporize 0.5 kg of water:

Q_v=m'\times h_{fg}

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Q_v=0.5\times 2256.4

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Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

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sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

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anyanavicka [17]

Answer:

The answer to your question is 5.4 cm

Explanation:

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Data

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Formula

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Substitution

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Simplification

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3 years ago
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