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Nadya [2.5K]
3 years ago
6

In the reactions,A products with the concentration [A]0=1.512M,[A]is found to be 1.496at t=30S.With the initial concentration [A

]0=2.584M,[A]is found to be 2.552M at t=1minute.What's the order of
Chemistry
1 answer:
Contact [7]3 years ago
8 0

Answer:

Zero order

Explanation:

Reaction rate for experiment 1 =

1.496 - 1.512 / 30 - 0 = - 5.3 * 10^-4 Ms-1

Reaction rate for experiment 2

2.552 - 2.584/60 - 0 =  - 5.3 * 10^-4 Ms-1

So,

Since the rate of reaction does not change with change in concentration of reactants (identified by the fact that the rates of reaction 1 and reaction 2 are the same), the reaction is zero order.

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Draw a resonance structure, complete with all formal charges and lone (unshared) electron pairs, that shows the resonance intera
Angelina_Jolie [31]

Answer:

See explanation

Explanation:

The acetoxy group in ortho position in phenyl acetate does interact with the phenyl moiety in the molecule via resonance.

This detailed interaction of the  acetoxy group in ortho position in phenyl acetate with the phenyl moiety in the molecule via resonance is shown in the image attached.

This interaction is made possible because the oxygen atom of the acetoxy group has lone pairs of electrons that are suitably positioned to interact with the ring via resonance.

8 0
3 years ago
The decomposition reaction of carbon disulfide to carbon monosulfide and sulfur is first order with k = 2.80 ✕ ✕ 10−7 sec-1 at 1
madam [21]

Answer: a) 1.97 grams of carbon disulfide will remain after 37.0 days.

              b) 2.85 grams of carbon monosulfide will be formed after 37.0 days.

Explanation: The decomposition of carbon disulfide is given as:

                          CS_2(g)\rightarrow CS(g)+S(g)

at t=0                    4.83g             0          0

at t=37 days        4.83 - x            x           x

here,

x = amount of CS_2 utilised in the reaction

This reaction follows first order kinetics so the rate law equation is:

k=\frac{2.303}{t}log\frac{A_o}{A}

where, k = rate constant

t = time

A_o = Initial mass of reactant

A = Final mass of reactant

a) For this, the value of

k=2.80\times10^{-7}sec^{-1}

t = 370 days = 3196800 sec

A_o = 4.83

A = 4.83-x

Putting values in the above equation, we get

2.8\times 10^{-7}sec^{-1}=\frac{2.303}{3196800sec}log\left(\frac{4.83}{4.83-x}\right)

x = 2.85g

Amount of CS_2 remained after 37 days = 4.83 - x

                                                                     = 1.97g

b) Amount of carbon monosulfide formed will be equal to "x" only which we have calculated in the previous part.

Amount of carbon monosulfide formed = 2.85g

6 0
3 years ago
The density of air under ordinary conditions at 25 degrees * C is 1.19g / L . How many kilograms of air are in a room that measu
professor190 [17]

Answer:

33.3 kg of air

Explanation:

This is a problem of conversion unit.

Density is mass / volume

Therefore we have to calculate the volume in the room, to be multiply by density. That answer will be the mass of air.

Volume of the room → 9 ft . 11 ft . 10 ft = 990 ft³

Density is in g/L, therefore we have to convert the ft³ to dm³ (1 dm³ = 1L)

990 ft³ . 28.3 dm³ / 1ft³ = 28017 dm³ → 28017 L

This is the volume of the room, if we replace it in the density formula we can know the mass of air in g.

1.19 g/L = Mass of air / 28017 L

Mass of air = 28017 L .  1.19 g/L → 33340 g of air

Finally, let's convert the mass in g to kg → 33340 g . 1kg / 1000 g = 33.3 kg

5 0
4 years ago
The distance between a carbon atom and an oxygen atom in the co molecule is how far from the carbon atom is the center of mass o
DENIUS [597]

The distance between a carbon atom and an oxygen atom in the co molecule is 0.63

<h3>What is the distance?</h3>

Distance is the measurement of the farness of two objects from each other.

The distance between the molecules can be calculated by the given formula

xcm = m_1+m_2m_1\times 1+m_2 \times 2

Putting the values in the equation

xcm = 12+1612 \times 0+16 \times 1.1 = 0.63 A^\circ

Thus, the distance between a carbon atom and an oxygen atom in the co molecule is 0.63

Learn more about distance

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5 0
2 years ago
Read 2 more answers
a nurse practitioner orders Medrol to be given 1.5 mg/kg of body weight. if a child weighs 72.6lb and the available stock of Med
Harrizon [31]
Answer is: 2,469 mL give to the child.
The mass m in kilograms (kg) is equal to the mass m in pounds (lb) times 0,45359237: m(child) = 72,6 · 0,045359237 = 32,93 kg.
m(Medrol) = 32,93 kg · 1,5 mg/kg.
m(Medrol) = 49,39 mg.
d(Medrol) = 20,0 mg/mL.
V(Medrol) = m(Medrol) ÷ d(Medrol).
V(Medrol) = 49,39 mg ÷ 20 mg/mL.
V(Medrol) = 2,469 mL.

8 0
3 years ago
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