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Salsk061 [2.6K]
3 years ago
13

Using this equation, m1v2=m2v2 , calculate the diluted molarity of 100 mL of a 0.5 M solution when 50 mL of

Chemistry
1 answer:
geniusboy [140]3 years ago
8 0

The molarity of the diluted solution is 0.33 M

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 0. 5 M

Volume of stock solution (V₁) = 100 mL

Volume of diluted solution (V₂) = 100 + 50 = 150 mL

<h3>Molarity of diluted solution (M₂) =? </h3>

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

<h3>M₁V₁ = M₂V₂</h3>

0.5 × 100 = M₂ × 150

50 = M₂ × 150

Divide both side by 150

M₂ = 50 / 150

<h3>M₂ = 0.33 M</h3>

Therefore, the molarity of the diluted solution is 0.33 M

Learn more: brainly.com/question/24625656

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Answer: The correct answer is option(A).

Explanation:

Total forces exerting on the car = F

F_r=50 N= Force on car exerting in right direction

F_l=20 N= Force on car exerting in left direction

F_u=2500 N= Force on car exerting in upward direction

F_d=2500 N= Force on car exerting in downward direction

F=F_r+F_l+F_u+F_d

(-F_u)=F_d (negative sign shows the direction)

Since, upward force are equal in magnitude but opposite in direction by which they will balance out each other.so, the net force car will be due to two forces F_r and F_l

F=F_r+F_l=50 N+(-20 N)=30 N(negative shows the direction)

The magnitude (size) and direction of the cumulative force acting on the car will 30 N towards right direction.Hence, correct answer is option(A).

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cupoosta [38]
The answer is:  " 1.75 * 10 ^(-10)  m " .
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Explanation: 
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This very question asked for "Question Number 3 (THREE) ONLY, which is fine!
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Given: " 0.000000000175 m " ;  write this in "scientific notation.
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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"
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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>
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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" .
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Then we write:  "  * 10^ "
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   {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:

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

And count the number of decimal spaces "backward" from the
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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" .
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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" .} . 
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The answer is:  " 1.75 * 10^(-10)   m " .
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