Answer: increase in thermal heat
explanation: In order to change the state, more energy is needed.
When you want to go from a lower state (solid) to a higher state (liquid) then you need more energy. and more energy means more heat in this case. if it was going from liquid to solid, you would need less thermal energy.
A critical mass is the smallest amount of fissile material needed for a sustained nuclear chain reaction.
75:15:10. componets inlcude <span>mixture of </span>sulfur<span>, charcoal, and </span>potassium nitrate<span> (</span>saltpeter<span>)</span>
<u>Answer:</u> The average atomic mass of copper is 63.55 amu.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
- <u>For
isotope:</u>
Mass of isotope = 62.94 amu
Percentage abundance of = 69.17 %
Fractional abundance of isotope = 0.6917
- <u>For
isotope:</u>
Mass of isotope = 64.93 amu
Percentage abundance of = 30.83 %
Fractional abundance of isotope = 0.3083
Putting values in equation 1, we get:
![\text{Average atomic mass of Copper}=[(62.94\times 0.6917)+(64.93\times 0.3083)]\\\\\text{Average atomic mass of copper}=63.55amu](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20Copper%7D%3D%5B%2862.94%5Ctimes%200.6917%29%2B%2864.93%5Ctimes%200.3083%29%5D%5C%5C%5C%5C%5Ctext%7BAverage%20atomic%20mass%20of%20copper%7D%3D63.55amu)
Hence, the average atomic mass of copper is 63.55 amu.
Stoichiometry:
First, calculate the number of grams for one mole of Ca3 (PO3)4
(3 * (Mass of Ca)) + (4 * (Mass of P + (3 * Mass of Oxygen)))
= (3*40.08) + 4(30.97 + (3*16.00))
=(120.24) + 4(78.97)
=436.12 g / mol Ca3(PO3)4
This means there are 436.12 g per 1 mole of Ca(PO3)4. Since there are 4.50 moles of Calcium Phosphate, mulitply the molar mass of Ca(PO3)4 by 4.50 and you should get 1962.54 g. Since there are 3 sigfigs, the final answer is 1960 g.
on a side note: I put in all my work in case 1. your periodic table if different, 2. my work is wrong, 3. you put in the question wrong because I feel that the actual compound would be Ca3(PO4)3 instead of Ca3(PO3)4 (if this is the case, the answer should be 1820 g).
