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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Answer:
Option D. 13.44
Explanation:
We'll begin by calculating the number of mole in 47.7g of copper(II) oxide, CuO.
This can be obtained as follow:
Mass of CuO = 47.7 g
Molar mass of CuO = 63.5 + 16 = 79.5 g/mol
Mole of CuO =.?
Mole = mass /Molar mass
Mole of CuO = 47.7/79.5
Mole of CuO = 0.6 mole
Next, we shall write the balanced equation for the reaction. This is given below:
CuO + H2 —> Cu + H2O
From the balanced equation above,
1 mole of CuO reacted with 1 mole of H2 to produce 1 mole of Cu and 1 mole of H2O.
Next, we shall determine the number of mole of H2 needed to react completely with 0.6 mole of CuO.
This can be obtained as follow:
From the balanced equation above,
1 mole of CuO reacted with 1 mole of H2.
Therefore, 0.6 mole of CuO will also react with 0.6 mole of H2.
Finally, we shall determine the volume occupied by 0.6 mole of H2 at STP.
This can be obtained as follow:
1 mole of H2 occupied 22.4 dm³ at STP.
Therefore, 0.6 mole of H2 will occupy = 0.6 × 22.4 = 13.44 dm³.
Therefore, 13.44 dm³ of H2 is needed for the reaction.
Answer:
Yes.
The nuclear equation {226/88 Ra → 222/26 Rn + 4/2 He} is balanced. As we know that an alpha particle is identical to a helium atom. This implies that if an alpha particle is eliminated from an atom's nucleus, an atomic number of 2 and a mass number of 4 is lost.
Therefore, the equation will be reduced to:
226 - 4 = 222
88 - 2 = 86
Hence, the equation is balanced.
Explanation:
Answer:
The Answer is D: A solid with no repeating pattern in its structure
Explanation:
