<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
A. The patch's area in square kilometers (km²) is 1.61×10⁻⁹ km²
B. The cost of the patch to the nearest cent is 734 cents
<h3>A. How to convert 16.1 cm² to square kilometers (km²)</h3>
We can convert 16.1 cm² to km² as illustrated below:
Conversion scale
1 cm² = 1×10⁻¹⁰ km²
Therefore,
16.1 cm² = 16.1 × 1×10⁻¹⁰
16.1 cm² = 1.61×10⁻⁹ km²
Thus, 16.1 cm² is equivalent to 1.61×10⁻⁹ km²
<h3>B. How to determine the cost in cent</h3>
We'll begin by converting 16.1 cm² to in². This can be obtained as illustrated below:
1 cm² = 0.155 in²
Therefore,
16.1 cm² = 16.1 × 0.155
16.1 cm² = 2.4955 in²
Finally, we shall the determine the cost in centas fo r llow:
- Cost per in² = $2.94 = 294 cent
- Cost of 2.4955 in² =?
1 in² = 294 cent
Therefore,
2.4955 in² = 2.4955 × 294
2.4955 in² = 734 cents
Thus, the cost of the patch is 734 cents
Learn more about conversion:
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Answer:
The amount of movement of tectonic plates in the area
Explanation:
The population density vary across the countries
Japan(127m), Philippines ( 103m), Indonesia (267m).
The volcanoes in Indonesia are among the most active of the pacific Ring of Fire along the north east island
Electronnegativity increase because the number of charges on the nucleus increases. Which attracts the bonding amount of electrons more.
