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Sergeeva-Olga [200]
3 years ago
14

A first-order reaction (A → B) has a half-life of 25 minutes. If the initial concentration of A is 0.900 M, what is the concentr

ation of B after 50 minutes?
Chemistry
1 answer:
Oksanka [162]3 years ago
7 0

Answer:

0.6749 M is the concentration of B after 50 minutes.

Explanation:

A → B

Half life of the reaction = t_{1/2}=25 minutes

Rate constant of the reaction = k

For first order reaction, half life and half life are related by:

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{25 min}=0.02772 min^{-1}

Initial concentration of A = [A]_o=0.900 M

Final concentration of A after 50 minutes = [A]=?

t = 50 minute

[A]=[A]_o\times e^{-kt}

[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}

[A] = 0.2251 M

The concentration of A after 50 minutes = 0.2251 M

The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M

0.6749 M is the concentration of B after 50 minutes.

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2. Determine the molarity of the NaOH solution in each trial. a. Trial 1 Molarity: b. Trial 2 Molarity: 3. Calculate the average
butalik [34]

Answer:

This question is incomplete

Explanation:

This question is incomplete but...

1) You can calculate the molarity of the NaOH for each trial by following the steps below.

The formula for Molarity (M) is

M = number of moles (n) ÷ volume (V)

where the unit of volume must be in Litres or dm³

The unit of molarity is mol/dm³ or mol/L or molar conc (M)

The final answer must have the unit of molarity

If the number of moles is not provided, look out for the mass of NaOH used and then calculate your number of moles (n) as

n = mass of NaOH used ÷ molar mass of NaOH

Where the atomic mass of sodium (Na) is 23, oxygen (O) is 16 and hydrogen (H) is 1. Hence, molar mass for NaOH is 23 + 16 + 1 = 40 g/mol

n = mass of NaOH used ÷ 40

2) Average Molarity will be (Trial 1 Molarity +Trial 2 Molarity) ÷ 2

Answer must be in mol/dm³ or mol/L or M

3) Label the volumentric flask containing the NaOH solution with the answer gotten from (2) above

3 0
3 years ago
Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
ElenaW [278]

Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

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calculate the enthalpy change for converting 1.00 mol of liquid water at 100 C to water vapor at 145 degree celsius for water Δ
Svetllana [295]

Answer:

q1 = mCpΔT

    = 18.016g × 1.84J/g.K × (418.15-373.15)

     = 1491.72 J

q2 = n×ΔH vap

     =  1mol ×44.0kJ/mol

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∴ qtotal = q1+ q2

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             = 45.498KJ

Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.

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