The correct answer is <span>Antoine-Laurent de Lavoisier. Hope this helps!</span>
Answer:
Mass = 255 g
Explanation:
Given data:
Number of moles of nitrogen = 7.5 mol
Mass of ammonia formed = ?
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
Now we will compare the moles of nitrogen and ammonia.
N₂ : NH₃
1 : 2
7.5 : 2/1×7.5 = 15
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 15 mol × 17 g/mol
Mass = 255 g
In this case, we are going to assume that there are 100 atoms to make things easier.
Let R% be the abundance of n-15. With this in mind, we calculate the abundance of n-14 to be 100%-R%
14.0031*(100-R)% + 15.001 * R%= 14.00674
In this case, we can delete or ignore the % sign since we do not want to carry it around, however, we need to keep in mind that the final answer is in %
14.0031*(100-R) + 15.001 * R= 14.00674
1400.31-14.0031R+15.001R=1400.674
0.9979R=0.364
R=0.3648
Then, the abundance of n-15 is 0.3648%
Converting the temperature, 295 K from Kelvin to Celsius scale:
Water has a boiling point of 
and a melting point of 
When we compare water at two different temperatures, 
we can say that water is in liquid form at both these temperatures as both of them are quite below the boiling temperature and above the melting temperature.
The temperature difference between water at the given two temperatures = 
Water at 
is at a higher temperature and so is warmer than water at a lower temperature of 
.
Answer: 733 kcal
Explanation:
According to a system used to calculate the available energy of foods, we have the following data:
- protein has 4 kcal/g
- carbohydrates has 4 kcal/g
- fat has 9 kcal/g
- alcohol has 7 kcal/g
So, we are told Jhon and his wife's dinner has:
- 55 g of carbohydrates
- 36 g of protein
- 27 g of fat
- 18 g of
Now we have to multiply these values by the kilocalories:
- (55 g of carbohydrates)(4 kcal/g)=220 kcal
- (36 g of protein)(4 kcal/g)=144 kcal
- (27 g of fat)(9 kcal/g)=243 kcal
- (18 g of alcohol)(7 kcal/g)=126 kcal
Then we have to sum all, giving the following result:
<h2>733 kcal</h2>
