Answer : The mass defect required to release energy is 6111.111 kg
Explanation :
To calculate the mass defect for given energy released, we use Einstein's equation:
E = Energy released = 
= mass change = ?
c = speed of light = 
Now put all the given values in above equation, we get:
Therefore, the mass defect required to release energy is 6111.111 kg
The fraction of the original amount remaining is closest to 1/128
<h3>Determination of the number of half-lives</h3>
- Half-life (t½) = 4 days
- Time (t) = 4 weeks = 4 × 7 = 28 days
- Number of half-lives (n) =?
n = t / t½
n = 28 / 4
n = 7
<h3>How to determine the amount remaining </h3>
- Original amount (N₀) = 100 g
- Number of half-lives (n) = 7
- Amount remaining (N)=?
N = N₀ / 2ⁿ
N = 100 / 2⁷
N = 0.78125 g
<h3>How to determine the fraction remaining </h3>
- Original amount (N₀) = 100 g
- Amount remaining (N)= 0.78125 g
Fraction remaining = N / N₀
Fraction remaining = 0.78125 / 100
Fraction remaining = 1/128
Learn more about half life:
brainly.com/question/26374513
We can calculate the final temperature from this formula :
when Tf = (V1* T1) +(V2* T2) / (V1+ V2)
when V1 is the first volume of water = 5 L
and V2 is the second volume of water = 60 L
and T1 is the first temperature of water in Kelvin = 80 °C +273 = 353 K
and T2 is the second temperature of water in Kelvin = 30°C + 273= 303 K
and Tf is the final temperature of water in Kelvin
so, by substitution:
Tf = (5 L * 353 K ) + ( 60 L * 303 K) / ( 5 L + 60 L)
= 1765 + 18180 / 65 L
= 306 K
= 306 -273 = 33° C
Answer:
P₅O₁₂
<em>Explanation: </em>
Assume that you have 100 g of the compound.
Then you have 44.7 g P and 55.3 g O.
1. Calculate the <em>moles</em> of each atom
Moles of P = 44.7 × 1/30.97 = 1.443 mol Al
Moles of O = 55.3 × 1/16.00 = 3.456 mol O
2. Calculate the <em>molar ratios</em>.
P: 1.443/1.443 = 1
O: 3.456/1.443 = 2.395
3. Multiply by a number to make the ratio close to an integer
P: 5 × 1 = 5
O: 5 × 2.395 = 11.97
3. Determine the <em>empirical formula
</em>
Round off all numbers to the closest integer.
P: 5
O: 12
The empirical formula is <em>P₅O₁₂</em>.
