You can use Le Chatelier's Principle to describe the equilibrium shift.
Le Chaterlier's Principle states that: "<span>If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change."
Thus, if you heat up the reaction, the equilibrium shift favors the endothermic reaction. If you increase pressure (if gases are involved), the shift favors the reaction that produces less gaseous products (to counteract pressure) and so on.</span>
Depends on the situation. If the nucleophile is already in excess, then no the reaction will not occur faster
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:
"1.75" .
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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).
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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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The [OH⁻] of the solution is 5.37×10⁵ M.
<h3 /><h3>What is pOH?</h3>
This is the negative logarithm to base 10 of hydroxy ion [OH⁻] concentration.
To calculate the hydroxy ion [OH⁻] concentration we use the formula below.
Note:
- pOH = 14-pH
- pOH = 14-9.77
- pOH = 4.27
Formula:
- [OH⁻] = 1/
................. Equation 1
Given:
Substitute the value into equation 1
- [OH⁻] = 1/
- [OH⁻] = 5.37×10⁵
Hence, The [OH⁻] of the solution is 5.37×10⁵ M.
Learn more about hydroxy ion concentration here: brainly.com/question/17090407
