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agasfer [191]
3 years ago
14

In the reaction __H2 + O2 → 2H2O, what coefficient should be placed in front of H2 to balance the reaction?

Computers and Technology
2 answers:
malfutka [58]3 years ago
8 0
The answer is:  [B]:  " 2 " .
_____________________________________________
Explanation:
____________________________________
Given the chemical equation:
____________________________________

     <u> ? </u> H₂ + O₂ → 2 H₂<span>O   ;
____________________________________________
</span>      → <span> We are asked, "What coefficient, if any — should be put in front of the:  " H</span>₂ " ;
__________________________________________________
          (which is on the "left-hand side" side of the chemical equation given—      the "reactants") ;

     →  to get a balanced chemical equation?
__________________________________________________
 →  Let us examine the "right-hand side" of this chemical equation—the product(s).  In this case, the "product" given is:  " 2 H₂O " .  

So, on the "right hand side", we have:
__________________________________________________
     1)  4 "H's" → {Two "H₂" 's = 2 * 2 = "4 H's"} ;  <u><em>and</em></u>:
     2)  2 "O's" → { Two "O's").
__________________________________________________
   So, the left-hand side should have:
__________________________________________________
     1)  4 "H's" ;  <u><em>and</em></u>:
     2)  2 "O's" ;
__________________________________________________
Let us examine the left-hand side (the "reactants").
__________________________________________________
                  " <u> ? </u><u /> H₂  +  O₂ " ; 
__________________________________________________
On the left-hand side, we already have: 
__________________________________________________
     1)   " 2  O's " ;   →  one "O₂" = "2 O's" ;  and:
     2)    "2  H's " ;   →  one "H<span>₂" .</span>

Now, we would need:
__________________________________________________
A "<u>total of "4 H's</u>".  Is there any number we could put as a coefficient on the other reactant, which happens to be:  "H₂" ;  to make a <u><em>total of</em></u> "4 H's" ?          

         {Note: There are only these TWO (2) reactants in this chemical equation.}.  

        → The "H₂" ;  as it stands alone, is insufficient—since that would be only                                                                                         "2 H's".
 __________________________________________________
     →  Thus, we can rule out:  "Answer choices: [A] and [D]."  
___________________________________________________
<u>Note</u>:  
→  Choice [A]:  "1" ;  The coefficient, "1";  is generally not (never?) used; and basically would function as the same as:
          →  Choice: [D]:  "<span>no coefficient is needed".
</span>→  <span>Choices [A] & [D]: would leave us with only "2 H's" on the "reactants side" (i.e. "left-hand side of the equation"; and we need FOUR ("4 H's").
__________________________________________________
Since we are given:  "H</span>₂" ; what coefficient could we put in front of this to get:       "4 H's" ?  (4÷2 =2).  So we could put a "2" in front of the "H₂" ; to get: 
    "4 H's".  The coefficient, "2" , corresponds directly with: 
__________________________________________________
                                             → Answer choice:  [B]:  "2" .
_________________________________________________
   { <u>NOTE</u>:  The remaining answer choice, [C], which is, "3" ; is INCORRECT; since 3 "H₂'s" would be "6 H's" ; (since 3 * "2 H's" equal "6 H's") ; which is too many "H's" → We need <u>exactly</u> "4 H's".}.
_________________________________________________
   So, the correctly balanced equation is:

             2 H₂ + O₂ → 2 H₂O  ; 
_________________________________________________
     → The coefficient that goes before the "H₂" ; to make this chemical equation balanced, is:  "2" . 
_________________________________________________
The correct answer is:  [B]:  "2" .  

      → The coefficient that goes before the "H₂" is:  "2" .
_________________________________________________
         Hope this answer—and {lengthy} explanation—is of help!
_________________________________________________
zvonat [6]3 years ago
8 0

The answer is B (2).

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Answer:

3 bits

Explanation:

Given a 4- way set associative cache that has 64 blocks of 16 words.

Therefore, the number of sets cache has:

\frac{64}{4} = 16

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The number of cache blocks can be calculated as:

16kB/16 = 1024 bytes/16 byte\times 16 = 256 cache blocks

Now,

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2 years ago
This program is to compute and write to a file the GPA for student scores read from an input file. The point values of the grade
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Answer:

#Take input from the user

filename=input('Enter name of input file: ')

total_units=0#Total number of units for a student

total_score=0#Total score of a student

#Assign a score to each grade. I have reduced 0.33 for each grade. Change here if you need to

grade={'A':4.0,'A-':3.67,'B+':3.34,'B':3.01,'B-':2.68,'C+':2.35,'C':2.02,'C-':1.69,'D+':1.36,'D':1.03}

try:

with open(filename,"r") as input_file:#Open the file for input(reading)

output_file=open("GPA_output.txt","w") #Create and open GPA_output.txt if it doesn't exist for writing

for student_record in input_file:#read from input file

if("," in student_record):#if there is a , that means this is the student name

student_name=student_record.strip('\n')#Remove \n from student name

continue

else:

if(" " in student_record):#If a line contains spaces then its the student's grades for a course

student=student_record.split(" ")#Split to find the coursename, units and grade

#student[0]=coursename student[1]=units student[2]=grade

total_units+=int(student[1])#Calculate total units for 1 student

total_score+=int(student[1])*grade[student[2].strip('\n')]#Find the total score of a student

#grade[student[2]] will lookup for the score that we initialized earlier

#if student has a grade as A then this will look up as grade['A'] which will return 4

#Find the total score as product of this grade and units for this course

continue

else:

if(total_units>0):#Check if score has been calculated for a student earlier

print("in")

GPA=total_score/total_units#Calculate the GPA

output_file.write("%-26s%.2f\n" % (student_name, GPA))#Write the GPA and Student Name in the required format

total_units=0#reset the units

total_score=0#reset the score

if(total_units>0):#Essential for the case when the file doesn't end with a new line(Check if total_units is not 0) which means a record is pending

#to be written to the file

GPA=total_score/total_units#Calculate the GPA

output_file.write("%-26s%.2f\n" % (student_name, GPA))#Write the GPA and Student Name in the required format

total_units=0#reset the units

total_score=0#reset the score

input_file.close()#Close the input file

output_file.close()#Close the output file

except IOError as e:

print("Problem in Opening the required file")#Print a message if file cannot be opened

Explanation:

The program has been tested for all 3 scenarios mentioned i.e.

a. If the file ends without \n

b. If the file ends with 2 \n

c. If the file ends with 1 \n

PS: The Program will run irrespective of how it ends. Even if there are many \n at the end of file, the program will be fine.

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2. A host computer with an IP address of 172.16.29.155 using a subnet mask of 255.255.255.0 and a gateway of 172.16.29.1 pings a
ASHA 777 [7]

Answer:

From the host IP address and subnet mask, you can tell that both computers are on the same network. Host 172.16.29.155 will need to send an ARP request broadcast for 172.16.29.83 to get the MAC address and update its ARP table if the host 172.16.29.83 has not already done so.

Explanation:

<h2>ADDRESS RESOLUTION PROTOCOL (ARP)</h2>

Address Resolution Protocol (ARP) is a procedure for mapping a logical address known as the IP address to a physical address known as MAC (Media Access Control) address in a local area network. The job of the ARP is essentially to translate 32-bit addresses to 48-bit addresses and vice-versa when using IP Version 4 (IPv4) to communicate on the network.

Most of the computer programs/applications use logical address (IP address) to send/receive messages, however, the actual communication happens over the physical address (MAC address) this happens in the layer 2 of the OSI model. An ARP request is a broadcast, and an ARP response is a Unicast.

Scenarios when ARP is used:

Scenario-1: The sender is a host and wants to send a packet to another host on the same network.

·        Use ARP to find another host’s physical address

Scenario -2: The sender is a host and wants to send a packet to another host on another network.

·        Sender looks at its routing table.

·        Find the IP address of the next-hop (router) for this destination.

·        Use ARP to find the router’s physical address

Scenario -3: the sender is a router and received a datagram destined for a host on another network.

·        Router checks its routing table.

·        Find the IP address of the next router.

·        Use ARP to find the next router’s physical address.

Scenario -4: The sender is a router that has received a datagram destined for a host in the same network.

·        Use ARP to find this host’s physical address.

The situation in the question above falls under scenario-1. From the host IP address and subnet mask, you can tell that both computers are on the same network. Host 172.16.29.155 will need to send an ARP request broadcast for 172.16.29.83 so to get the MAC address and update its ARP table if the host 172.16.29.83 has not already done so.

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