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CaHeK987 [17]
3 years ago
5

Urgentttt help will be much appreciated

Chemistry
1 answer:
baherus [9]3 years ago
8 0

Answer:

A I guess not sure hope it helps

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The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
Explain vapour pressure and surface tension What is plasma
worty [1.4K]

Vapor Pressure: Measures how a substance is likey to evaporate

Surface Tension: is the attraction between liquid molecules.

Plasma: is an organic and inorganic substance that is typically found in blood

5 0
4 years ago
The molar mass of O2 is 32.0 g/mol.what mass,in grams,of O2 is required to react completely with 4.00 mol of Mg
Dahasolnce [82]
The balanced equation for the reaction between Mg and O₂ is as follows
2Mg + O₂ --> 2MgO
stoichiometry between Mg and O₂ is 2:1
number of Mg reacted - 4.00 mol
if 2 mol of Mg reacts with 1 mol of O₂
then 4.00 mol of Mg requires - 1/2 x 4.00 = 2.00 mol of O₂
then the mass of O₂ required - 2.00 mol x 32.0 g/mol = 64.0 g
64.0 g of O₂ is required for the reaction 
7 0
4 years ago
Read 2 more answers
OhDjdkdkdkdrkfkfkfkfkfkfkffififkfktkt
Sergio039 [100]

Answer:

2 is the answer

Explanation:

Si: 1s2 2s2 2p6 3s2 3p2

Number of the unpaired electron will be 2

8 0
3 years ago
!!please help asap ty!!
Bess [88]

Answer:

2.89× 10²³ atom

Explanation:

Given data:

Number of atoms of Zn = ?

Number of moles of Zn = 0.48 mol

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atom

0.48 mol ×  6.022 × 10²³ atom / 1mol

2.89× 10²³ atom

7 0
3 years ago
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