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Simora [160]
3 years ago
8

If 7.84 × 107 J of energy is released from a fusion reaction, what amount of mass in kilograms would be lost? Recall that c = 3

× 108 m/s.
7.06 × 1024



8.71 × 10-4



8.71 × 10-7



8.71 × 10–10
Chemistry
2 answers:
IrinaK [193]3 years ago
8 0

Answer:

The amount of mass in kilograms would be lost 8.7111\times 10^{-10} kg.

Explanation:

Energy released during fusion reaction = 7.84\times 10^7 J

Mass of amount lost during the reaction = Δm

Speed of the light = c = 3\times 10^8 m/s

Using Einstein equation :

E=\Delta m\times c^2

7.84\times 10^7 J=\Delta m\times (3\times 10^8 m/s)^2

\Delta m=8.7111\times 10^{-10} kg

The amount of mass in kilograms would be lost 8.7111\times 10^{-10} kg.

katrin [286]3 years ago
3 0
  

m = 7.84x107/(3x108)2kg  = 7.84x107/9x1016kg = 0.871x10-9 kg = 8.71x10-10 kg





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How many grams are in 3.14 moles of PI₃?
OverLord2011 [107]

Answer:

\boxed {\boxed {\sf 1290 \ g \ PI_3}}

Explanation:

We want to convert from moles to grams, so we must use the molar mass.

<h3>1. Molar Mass</h3>

The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.

  • Phosphorus (P): 30.973762 g/mol
  • Iodine (I): 126.9045 g/mol

Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.

  • I₃: 126.9045 * 3=380.7135 g/mol
  • PI₃: 30.973762 + 380.7135 = 411.687262 g/mol

<h3>2. Convert Moles to Grams</h3>

Use the molar mass as a ratio.

\frac {411.687262 \ g \ PI_3}{ 1 \  mol \ PI_3}

We want to convert 3.14 moles to grams, so we multiply by that value.

3.14 \ mol \ PI_3 *\frac {411.687262 \ g \ PI_3}{ 1 \  mol \ PI_3}

The units of moles of PI₃ cancel.

3.14 *\frac {411.687262 \ g \ PI_3}{ 1 }

1292.698 \ g\ PI_3

<h3>3. Round</h3>

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.

  • 1292.698

The 2 in the ones place tells us to leave the 9.

1290 \ g \ PI_3

3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>

4 0
3 years ago
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