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Fudgin [204]
2 years ago
12

Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ

ine whether the reaction is first order or second order. determine whether the reaction is first order or second order. this is the first order reaction. this is the second order reaction. submitprevious answers correct part b what is the rate constant?
Chemistry
1 answer:
Whitepunk [10]2 years ago
7 0
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
   = 8.25 x 10⁻³ s⁻¹

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A 23.0g sample of a compound contains 12.0g of C, 3.0g of H, and 8.0g of O.What the empirical formula of the compound
Kryger [21]

Answer:

The empirical formula of compound is C₂H₆O.

Explanation:

Given data:

Mass of carbon = 12 g

Mass of hydrogen = 3 g

Mass of oxygen = 8 g

Empirical formula of compound = ?

Solution:

First of all we will calculate the gram atom of each elements.

no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms

no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms

no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms

Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.

          C:H:O  =     1/0.5  :   3/0.5  :   0.5/0.5

          C:H:O  =     2      :     6      :     1

The empirical formula of compound will be C₂H₆O

5 0
2 years ago
How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge
Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:-  The balanced equation for the combustion of ethane is:

2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})

= 126 L O_2

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

105.0LO_2(\frac{4LCO_2}{7L O_2})

= 60.0 L CO_2

Similarly, let's calculate the volume of water vapors formed:

105.0L O_2(\frac{6L H_2O}{7L O_2})

= 90.0 L H_2O

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

105.0LO_2(\frac{2LC_2H_6}{7LO_2})

= 30.0 L C_2H_6

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

5 0
2 years ago
2.A solid block with a length of 6.0 cm, a width of 3.0 cm, and a height of 3.0 cm has a mass of 146 g. What is the block’s dens
posledela
2: <span>Volume V = a*b*c = 6.0*3.0*3.0 = 54.0 cm^3                                                   density ρ = mass/volume = 146/54 = 2.70 g/cm^3
3: Volume = (27.8 -21.2) cm^3
mass = 22.4 g
density = 22.4/(27.8-21.2) g/cm^3

</span>
3 0
2 years ago
Read 2 more answers
Calculate the number of chloride ions in 6.8 g of zinc chloride
Bumek [7]

Answer:

The correct answer is 0.300 * 10^23 ions.

Explanation:

Based on the given question, there is a need to find the number of chloride ions in the mentioned 6.8 grams of zinc chloride compound.  

The moles of zinc chloride (ZnCl2) is,  

= mass of zinc + 2 mass of chlorine

= 65.38 + 2 (35.45)

=65.38 + 70.90

= 136.28 grams (The molecular mass of zinc is 65.38 and the molecular mass of chlorine is 35.45)

Thus, 136.28 g of ZnCl2 contains 70.90 grams of chlorine

Therefore, 6.8 grams of ZnCl2 will comprise = (70.90/136.28) * 6.8

= 3.537 g of chlorine

70.90 g of Cl comprise 6.022*10^23 chlorine, thus, 3.537 g of Cl will comprise (6.022*10^23/70.90) * 3.537

= 0.300 * 10^23 ions of chlorine.  

3 0
2 years ago
Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base
Vesna [10]

Answer : The pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

Explanation :

First we have to calculate the value of K_b.

As we know that,

K_a\times K_b=K_w

where,

K_a = dissociation constant of an acid = 1.8\times 10^{-5}

K_b = dissociation constant of a base = ?

K_w = dissociation constant of water = 1\times 10^{-14}

Now put all the given values in the above expression, we get the dissociation constant of a base.

1.8\times 10^{-5}\times K_b=1\times 10^{-14}

K_b=5.5\times 10^{-10}

Now we have to calculate the concentration of hydroxide ion.

Formula used :

[OH^-]=(K_b\times C)^{\frac{1}{2}}

where,

C is the concentration of solution.

Now put all the given values in this formula, we get:

[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}

[OH^-]=1.3\times 10^{-5}M

Now we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (1.3\times 10^{-5})

pOH=4.9

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1

Therefore, the pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

4 0
2 years ago
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