The formula of alcohol are
1.913 g CO2/44 = 0.043 moles of CO2*12 = 0.043 moles of C 1.174 gram H2O/18 = 0.065 moles of H.
O: 1 - 0.516 g C - 13 g H = 0.354 g O/16 = 0.022 moles of O.
we have 0.043:0.130.;0.022 as C:H:O. Diving by the smallest no. 0.022 we get 1.95:5.9:1 or 2:6:1 as C:H:O.
hope this help
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:
c) .51835
Explanation:
Let the relative abundance of the lighter of the two isotopes be X we have
Then the relative abundance of the heavier isotope is then (1-X)
Whereby we have that in nature the amount of the lighter silver found in proportion is X and the heavier isotope of silver is present as (1-X) proportion in nature.
To calculate the relative atomic mass of silver, we have
(Mass of light weight silver)×X + (mass of heavier isotope of silver×(1-X) = relative atomic mass of silver
106.90509(X) + 108.9047(1-X)
108.9-108.9(x)+106.9(x) = 107.87
-2x-1.03 = 0.517450902926
Closest answer is c
c) .5184
The relative atomic mass of isotopes is the weighted average by the mole-fraction of abundance of these isotopes which gives the atomic weight that is listed for that element on the periodic table.
Answer : The precipitate will not be formed when these solutions are mixed.
Explanation :
The chemical equation for the reaction of calcium chloride and lithium sulfate follows:
We are given:
Concentration of calcium ion =
Concentration of sulfate ion =
The salt produced is calcium sulfate.
The equation follows:
The expression of (ionic product) for above equation follows:
Putting values of the concentrations in above expression, we get:
There are 3 conditions:
- When ; the reaction is product favored. (No precipitation)
- When ; the reaction is reactant favored. (Precipitation)
- When ; the reaction is in equilibrium. (Sparingly soluble)
As, the . The above reaction is product favored. This means that no salt or precipitate will be formed.
Hence, the precipitate will not be formed when these solutions are mixed.