Answer:
A) It's correctly written
B) 77%
C) 835 calories
Explanation:
A) From online sources, we have number of calories as follows;
Fats: 9 calories per gram
Protein; 4 calories per gram
Carbs; 4 calories per gram
Total calories for each;
Total fat = 3 × 9 = 27 calories
Total protein = 3 × 4 = 12 calories
Total carbs = 32 × 4 = 128 calories
(sugar and dietary Fibre are classified as carbohydrates and so total carbs takes care of their calories).
Thus, total number of calories per serving = 27 + 12 + 128 = 167 calories per serving which is same as what is given.
B) percent from carbohydrates per serving = total calories from carbs/total number of calories per serving × 100% = 128/167 × 100% ≈ 77%
C) One box contains 5 servings. Thus total number of calories per box = 167 × 5 = 835 calories
Wind ,ice,and water eroded it over years
The masses of the objects and the distance between them
-hope it helps
Answer:
1. 4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. 6 moles of Cl2
Explanation:
1. The balanced equation for the reaction. This is illustrated below:
4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. Determination of the number of mole of Cl2 produce when 4 moles of FeCl3 react with 4 moles. To obtain the number of mole of Cl2 produced, we must determine which reactant is the limiting reactant.
This is illustrated below:
From the balanced equation above,
4 moles of FeCl3 reacted with 3 moles of O2.
Since lesser amount of O2 (i.e 3 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of FeCl3, therefore FeCl3 is the limiting reactant and O2 is the excess reactant.
Finally, we can obtain the number of mole Cl2 produced from the reaction as follow:
Note: the limiting reactant is used as it will produce the maximum yield of the reaction since all of it is used up in the reaction.
From the balanced equation above,
4 moles of FeCl3 will react to produced 6 moles of Cl2.
Answer:
Explanation:
From the given information:
The concentration of metal ions are:
![[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D%20%5Cdfrac%7B0.003474%20%5C%20M%20%5Ctimes%2020.49%20%5C%20mL%7D%7B10.0%20%5C%20mL%7D)
![[Ca^{2+}]=0.007118 \ M](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D0.007118%20%5C%20M)
![[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BMg%5E2%2B%5D%20%3D%20%5Cdfrac%7B0.003474%20%5C%20M%5Ctimes%20%2826.23%20%20-%2020.49%20%29mL%7D%7B10.0%20%5C%20mL%7D)
Mass of Ca²⁺ in 2.00 L urine sample is:
= 0.1598 g
Mass of Ca²⁺ = 159.0 mg
Mass of Mg²⁺ in 2.00 L urine sample is:
= 0.3461 g
Mass of Mg²⁺ = 346.1 mg
