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:
In comparison to Part 1 of this experiment, we observed similar reactions when determining the make up of our unknown. When testing for Mn2+ we observed a color change that resulted in a darker brown/red color, when testing for Co2+ we observed the formation of foamy bubbles but we could not conclude that a gas had formed, when testing for Fe3+ the result was a liquid red in color, when testing for Cr3+ we observed no change, when testing for Zn2+ we observed the formation of a pink/red liquid, when testing for K+ we observed the formation of a precipitate, when testing for Ca2+ we observe the formation of a precipitate. Sources of error may have occurred when observing whether or not an actual reaction had taken place or not, using glassware that wasn't fully cleaned, or the accidental mix of various other liquids in the lab
Explanation:
Answer:
C. 10.540 moles
Explanation:
divide grams by molar mass to get moles
Take 23mols and multiply it by the atomic mass of Li:
23mol * 6.94g/mol = 159.62g
