Answer:
The percentage of the daily value for saturated fat provided by one package of this food is 26.0%
Explanation:
It has been given from the preceding statement that one package of the dinner provides 5.2g out of the 20 grams recommended for daily need. This is 5.2/20 x 100 given 26.0% of the daily neded value will be provided by a plate or package of the dinner.
The masses can be found by substractions:
- Mass of CaSO₄.H2O (hydrate):
16.05 g - 13.56 g = 2.49 g
15.07 g - 13.56 g = 1.51 g
- The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:
2.49 g - 1.51 g = 0.98 g
- The percent of water is found by the formula:
massWater ÷ massHydrate * 100%
0.98 g ÷ 2.49 g * 100% = 39.36%
- The mole of water is calculated using water's molecular weight (18g/mol):
0.98 g ÷ 18 g/mol = 0.054 mol water
- A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)
1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄
- The ratio of mole of water to mole of anhydrate is:
0.054 mol water / 0.011 mol CaSO₄ = 0.49
In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O
Depending in the category of the Hurricane, you make experience different levels of wind power and destruction. Hurricanes only have 5 categories ranking from Category 1 to Category 5. The smallest category is category 1 making category 5 the largest. The bigger the category, the more wind or destruction you'll experience.
Answer:
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Explanation:
Hello,
In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:
![[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D2.0x10%5E%7B-3%7D%5Cfrac%7BmolCa%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B2molF%5E-%7D%7B1molCa%5E%7B2%2B%7D%7D%20%20%5C%5C)
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Best regards.
1 mole=6.02 x 10^23 atoms so how many moles are there in 3.0 x 10^23 we will cross multiply, 1 x 3.0 x 10^23 / 6.02 x 10 ^23. Which will give us 0.498 moles.
Hope this helped