Answer:
Explanation:
The formula for the heat released is
q = mCΔT
Data:
m = 395 g
C = 4.184 J·°C⁻¹g⁻¹
ΔT = 55 °C
Calculations:
q = 395 × 4.184 × 55 = 91 000 J = 91 kJ
The water will absorb 
of energy.
N2 + 3H2 --> 2NH3
When 100g of N2 , no of moles of N2= 100/(28)=3.57 mol
no. of moles of h2 = 6/(2)=3mol
Therefore h2 is limiting reagent.
no. of moles of ammonia= 3/3*2=2moles
mass of ammonia produced= 2 mol * (14+3)= 34g
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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Combustion reactions release energy in the form of heat.
Hope this helps and tell me if you need more help!
