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aalyn [17]
2 years ago
13

Identify the reaction that would possibly make a good

Chemistry
1 answer:
Rama09 [41]2 years ago
3 0

Answer:

D

Explanation:

Cuz i got it wrong and this was the right answer : )

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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
Whitepunk [10]
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

7 0
3 years ago
What is the molarity of an H3PO4solution if 15.0 mL is completely neutralized by 38.5 mL of 0.150 MNaOH? Please explain your ans
rodikova [14]

Answer:

M=Molarity

V=volume

change millilitre to litre

so:15ml=0.015 L

:38.5ml=0.0385 L

M1V1=M1V1

M1* 0.015 L= 0.15M*0.0385 L

M1=0.00385

8 0
3 years ago
Functional group and bond hybridization of vanillin
lana66690 [7]

Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group:  R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a  <u>sp³</u>  hybridization because they are involved only in single bonds.

7 0
3 years ago
I'm having trouble balancing these. Can you help me​
AlekseyPX

Answer:

Oh Okay

I hope u have a better rest of ur day

4 0
3 years ago
What is the missing number in the sequence 3,5,9,_,15,23,33
balu736 [363]

Answer:

In every sequence even numbers are added in order.

3, 5, 9, 15, 23, 33

6 0
3 years ago
Read 2 more answers
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