Lab 4
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Toronto Metropolitan University *
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Subject
Chemistry
Date
Jan 9, 2024
Type
docx
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Uploaded by JusticeStarWaterBuffalo329
Lab 4.py
# Sahib Gaba
# 500927758
class Node:
def __init__
(
self
, dataval=
None
):
self
.dataval = dataval
self
.nextval = None
self
.prevval = None
class LinkedList:
def __init__
(
self
):
self
.headval = None
def __str__
(
self
):
node = self
.headval
output = "[ "
while
(node != None
):
output = output + str
(node.dataval) + ", "
node = node.nextval
if
(
len
(output) > 2
):
output = output[:-
2
]
return output + " ]"
# DO NOT EDIT ANY CODE ABOVE THIS LINE
def insert(
self
, val):
node=Node(val)
if self
.headval is None
:
self
.headval=node
return
end=
self
.headval
while
(end.nextval):
end=end.nextval
end.nextval=node
# Fill in this method, which takes a value and adds a node which holds the value at the end of the linked list
def delete(
self
, index=
0
):
curr_node = self
.headval
pointerIndex=
0
if pointerIndex == index:
self
.headval=curr_node.nextval
curr_node = None
return
prev = None
while curr_node is not None and pointerIndex!= index:
pointerIndex+=
1
prev = curr_node
curr_node=curr_node.nextval
if curr_node is None
:
return
prev.nextval=curr_node.nextval
curr_node
=
None
# Fill in this method, which takes an index(with a default value of 0), and deletes the node at the index specified
def find(
self
, val):
curr_node=
self
.headval
pointerIndex=
0
if curr_node.dataval == val:
return pointerIndex
while curr_node is not None and curr_node.dataval != val:
pointerIndex+=
1
curr_node = curr_node.nextval
if curr_node is None
:
prev = curr_node
return False
return pointerIndex
# Fill in this method, which takes in a value and returns the index of the first node which contains that value. If no node containing that value is found, return False
def reverse(
self
):
curr_node = self
.headval
prev = None
while curr_node is not None
:
next=curr_node.nextval
curr_node.nextval=prev
prev=curr_node
curr_node=next
self
.headval=prev
# Fill in this method, which reverses the list
# DO NOT EDIT ANY CODE PAST THIS LINE
# Tests
# Insertion Test
a = LinkedList()
a.insert(
1
)
a.insert(
2
)
a.insert(
'a'
)
a.insert(
3
)
print
(
str
(a) == "[ 1, 2, a, 3 ]"
)
# Deletion
a.delete()
print
(
str
(a) == "[ 2, a, 3 ]"
)
a.delete(
2
)
print
(
str
(a) == "[ 2, a ]"
)
# Find
a.insert(
'b'
)
a.insert(
'c'
)
a.insert(
'b'
)
print
(a.find(
'b'
) == 2
)
print
(a.find(
'c'
) == 3
)
print
(a.find(
'd'
) == False
)
# Reverse
a.reverse()
print
(
str
(a) == "[ b, c, b, a, 2 ]"
)
# Singly Linked Lists Tests
class DoublyLinkedList:
def __init__
(
self
):
self
.headval = None
def __str__
(
self
):
node = self
.headval
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Molarity (M) of NaOH (from the bottle of NaOH): 0.204 mol/L
Titration Number
3
4
34.44
mL
0.50ML
Final Volume buret reading (mL NAOH)
34.00
33.85
mL
0.50mL
33.80
mL
0.50mL
0.50mL
Initial Volume buret reading (mL NaOH)
Volume NaOH used in titration (mL) =
Final Volume buret reading (mL NAOH) - mL
Initial Volume buret reading (mL NaOH)
33.94
For calculations multiply mL by 10-3 to
convert mL toL
Molarity (M or mol/L) NaOH
from the bottle of NaOH
0.204
mol/L
10.0 mL 10.0mL
0.204
0.204
0.204
mol/L
10.0 mL 10.0 mL
mol/L
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A chemist must prepare 275. mL of 325. mM aqueous sodium carbonate (Na,CO,) working solution. He'll do this by pouring out some 0.896
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O STOICHIOMETRY
Finding mole ratios from chemical formulae
This is the chemical formula for talc (the main ingredient in talcum powder):
Mg3 (Si,O3),(OH),
An analytical chemist has determined by measurements that there are 0.0837 moles of silicon in a sample of talc. How many moles of oxygen are in the
sample?
Round your answer to 3 significant digits.
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O ELECTROPHILIC ADDITION REACTIONS
Recognizing common alkene addition reactions
X
Classify each of the following organic reactions.
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reaction
1) Hg(OAc)₂, H₂O
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HBr
ROOR
H₂
~=~
Pd
Cl₂
Br
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CI
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OH
type of reaction
(check all that apply)
inhalation
hydrohalogenation
addition
Ohydration
Odihydroxylation
halohydrin formation
addition
Odihydroxylation
Ohydration
halohydrin formation
substitution
hydrohalogenation
O halohydrin formation
subtraction
addition
Ohydration
Odihydroxylation
Ohydrohalogenation
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Using a chemical equation to find moles of product from moles ...
Wine goes bad soon after opening because the ethanol (CH,CH,OH) dissolved in it reacts with oxygen (0,) gas to form water and aqueous acetic acid
(CH,COOH), the main ingredient in vinegar. Calculate the moles of oxygen needed to produce 0.055 mol of acetic acid. Be sure your answer has a unit
symbol, if necessary, and round it to the correct number of significant digits.
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