The number of electrons that can be held in the second orbit are 8
Answer:
32 g Cu
Explanation:
1 mol Cu -> 63.5 g
0.5 mol Cu ->x
x=(0.5 mol *63.5 g)/1 mol x= 32 g Cu
The answer is C because it says describe the final sugar and C is unsaturated and in the problem it says as much as sugar will dissovle .
84 m/s x 31.221 s = 2600 m
-rounded from 2622.564 m because 84 m/s has only two significant figures
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:
brainly.com/question/2139943
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