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mina [271]
3 years ago
12

Consider the single‑step, bimolecular reaction. CH3Br+NaOH⟶CH3OH+NaBr When the concentrations of CH3Br and NaOH are both 0.120 M

, the rate of the reaction is 0.0080 M/s. What is the rate of the reaction if the concentration of CH3Br is doubled? rate: M/s What is the rate of the reaction if the concentration of NaOH is halved? rate: M/s What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of 5? rate: M/s
Chemistry
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

Explanation:

CH₃Br+NaOH⟶CH₃OH+NaBr

It is a single step bimolecular reaction so  order of reaction is 2 , one for CH₃Br and one for NaOH .

rate of reaction = k x [CH₃Br] [ NaOH]

.008 = k x .12 x .12

k = .55555

when concentration of CH₃Br is doubled

rate of reaction = .555555 x [.24] [ .12 ]

= .016 M/s

when concentration of NaOH  is halved

rate of reaction = .555555 x [.12] [ .06 ]

= .004  M/s

when concentration of both CH₃Br and Na OH is made 5 times

rate of reaction = .555555 x .6 x .6

= 0.2 M/s

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Answer:

Wavelength of the first four spectral line:

\lambda_1=1216{\AA}

\lambda_2=1026{\AA}

\lambda_3=972.8{\AA}

\lambda_4=950{\AA}

Explanation:

Lman series when electron from higher shell number and jumps to 1st shell then all the possible lines is called as lyman series.

to calculate wavelength of first four spectral line:

For hydrogen Z=1;

by using rydberg equation

\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]

1. n=2 to n=1

\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]

\frac{1}{R} =912{\AA} =rydberg constant

\lambda=\frac{4}{3R}

\lambda_1=1216{\AA}

2. n=3 to n=1

\lambda=\frac{9}{8R}

\lambda_2=1026{\AA}

3. n=4 to n=1

\lambda=\frac{16}{15R}

\lambda_3=972.8{\AA}

4. n=5 to n=1

\lambda=\frac{25}{24R}

\lambda_4=950{\AA}

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(You do) If you have 47.2 mol of Na available, along with an excess of Cl₂, how many grams of NaCl can you produce?
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Answer:

2,760 grams NaCl

Explanation:

To find grams of NaCl, you need to (1) convert moles of Na to moles of NaCl (via mole-to-mole ratio from reaction) and (2) convert moles of NaCl to grams (via molar mass from periodic table). The final answer should have 3 significant figures based on the given measurement.

2 Na + Cl₂ --> 2 NaCl

Molar Mass (NaCl) = 22.99 g/mol + 35.45 g/mol

Molar Mass (NaCl) = 58.44 g/mol

47.2 moles Na           2 moles NaCl              58.44 grams

----------------------  x  ---------------------------  x  -------------------------  =
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= 2,758.368 grams NaCl

= 2,760 grams NaCl

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