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Nookie1986 [14]
4 years ago
15

Please help with 3! Please give only the correct answer...

Chemistry
1 answer:
cupoosta [38]4 years ago
4 0
The answer is:  " 1.75 * 10 ^(-10)  m " .
_________________________________________________________
Explanation: 
_________________________________________________________
This very question asked for "Question Number 3 (THREE) ONLY, which is fine!
_________________________________________________________
Given: " 0.000000000175 m " ;  write this in "scientific notation.
_________________________________________________________
Note:   After the "first zero and the decimal point" {Note: that first zero that PRECEDES the decimal point in merely a "placeholder" and does not count as a "digit" — for our purposes} —
                     There are NINE (9) zeros, followed by "175"
_______________________________________________________
To write in "scientific notation", we find the integer that is written, as well, as any "trailing zeros" (if there are any—and by "trailing zeros", this means any number consecutive zeros/and starting with "the consecutive zeros" only —whether forward (i.e., "zeros following"; or backward (i.e. "zeros preceding").

In our case we have "zeros preceding";  that is a decimal point with zeros PRECEDING an "integer expression"<span>
</span><span> (the "integer" is "175").</span>
______________________________________________________
We then take the "integer expression" (whatever it may be:  12, 5, 30000001 ; or could be a negative value,  etc.) ;  

→  In our case, the "integer expression" is:  "175" ;

and take the first digit (if the expression is negative, we take the negative value of that digit;  if there is only ONE digit (positive or negative), then that is the digit we take ;

And write a decimal point after that first digit (unless in some cases, there is only one digit);  and follow with the rest of the consecutive digits of that 'integer expression' ;

→ In our case:  "175" ; becomes:  " 1.75" .
__________________________________________________
Then we write:  "  * 10^ "
__________________________________________________
   {that is "[times]"; or "multiplied by" :    [10 raised exponentially to the power of  <u>     </u> ]._____________________________________________________
 And to find that power, we take the "rewritten integer value (i.e. "whole number value that as been rewritten to a single digit with a decimal point"); and count the [number of "trailing zeros";  if there are any; PLUS the number of decimal places one goes] ; and that number is the value to which "10" is raised.
{If there are none, we write:  " * 10⁰ " ;    since "any value, raised to the "zero power", equals "1" ; so " * 10⁰ " ; is like writing:  " * 1 " .

If there are "trailing zeros" AND/OR or  any number of decimal places,  to the "right" of this expression; the combined number of spaces to the right is: 
  { the numeric value (i.e. positive number) of the power to which "10" is raised }.

Likewise, if there are "trailing zeros" AND/OR or any number of decimal places, to the "LEFT" of this expression; the combined number of spaces to the LEFT is the value of the power which "10" is raised to; is that number—which is a negative value.

In our case:  we have:  0.000000000175 * 10^(-10) .

Note:  The original notation was:

             →  " 0.000000000175 m "

{that is:  "175" [with 9 (nine) zeros to the left].}.

We rewrite the "175" ("integer expression") as:

"1.75" .
____________________________________________________
So we have:
         →   " 0.000000000175 m " ;

Think of this value as:

        " 0. 0000000001{pseudo-decimal point}75   m ".

And count the number of decimal spaces "backward" from the
      "pseudo-decimal point" to the actual decimal; and you will see that there are "10" spaces (to the left).   
______________________________________________________
Also note:  We started with "9 (nine)" preceding "zeros" before the "1" ;  now we are considering the "1" as an "additional digit" ;
             →  "9 + 1 = 10" .
______________________________________________________
Since the decimals (and zeros) come BEFORE (precede) the "175" ; that is, to the "left" of the "175" ; the exponent to which the "10" is raised is:
 "NEGATIVE TEN" { "-10" } .

So we write this value as:  " 1.75 * 10^(-10)  m " .  

{NOTE:  Do not forget the units of measurement; which are "meters" —which can be abbreviateds as:  "m" .} . 
______________________________________________________
The answer is:  " 1.75 * 10^(-10)   m " .
______________________________________________________
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Which statement about electrons is not true?a. One electron state is a bonding orbital between nucleib. One electron state is an
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Answer:

b. One electron state is an anti-bonding orbital, which results in an absence of electron density between atoms.

Explanation:

8 0
4 years ago
in example 5.11 of the text the molar volume of n2 at STP is given as 22.42 L/mol how is this number calculatd how does the mola
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Answer:

V = 22.42 L/mol

N₂ and H₂ Same molar Volume at STP

Explanation:

Data Given:

molar volume of N₂ at STP = 22.42 L/mol

Calculation of molar volume of N₂ at STP  = ?

Comparison of molar volume of H₂ and N₂ = ?

Solution:

Molar Volume of Gas:

The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol

Molar volume can be calculated by using ideal gas formula  

                               PV = nRT

Rearrange the equation for Volume

                            V = nRT / P . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

Standard values

P = 1 atm

T = 273 K

n = 1 mole

R = 0.08206 L.atm / mol. K

Now put the value in formula (1) to calculate volume for 1 mole of N₂

                   V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm

                   V = 22.42 L/mol

Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.

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3 years ago
When mixed, solutions of silver nitrate, AgNO3, and sodium sulfate, Na2SO4, form a precipitate of silver sulfate, Ag2SO4. The ba
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Answer:

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

Explanation:

The general schemefor a reaction is given as;

Reactants --> Products

In this question, the reactants are AgNO3 and Na2SO4. The product is Ag2SO4.

The equation is given as;

AgNO3 + Na2SO4 --> Ag2SO4

The other poduct formed in this reaction is NaNO3.

The full reaction is given as;

AgNO3 + Na2SO4 --> Ag2SO4 + NaNO3

The above reaction is not balanced because there are unequal number of atoms of the elements on both sides of the reaction.

The balanced equation is given as;

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

In this equation, there are equal number of moles of the atoms on both sides.

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