Answer:
The answer is "6.52 kg and 13.1 kg"
Explanation:
For point a:
Equation:
Calculating the amount of
![= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg](https://tex.z-dn.net/?f=%3D%209.88%20%5C%20kg%20%5Ctimes%20%5Cfrac%7B1000%20%5C%20g%7D%7B1%20%5C%20kg%7D%20%5Ctimes%20%5Cfrac%7B1%20%5C%20mol%20%5C%20MnO_2%7D%7B86.94%20%5C%20g%7D%20%5Ctimes%20%5Cfrac%7B2%20%5C%20Mol%20%5C%20MnCO_3%7D%7B2%20%5C%20mol%20%5C%20MnO_2%7D%20%5Ctimes%20%5Cfrac%7B114.95%5C%20g%7D%7B1%20%5C%20mol%20%5C%20MnCO_3%20%7D%5Ctimes%20%5Cfrac%7B1%5C%20kg%7D%7B1000%5C%20g%7D%5C%5C%5C%5C%3D%20%2013.1%20%5C%20kg)
For point b:
![Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\](https://tex.z-dn.net/?f=Actual%5C%20yield%20%3D%204.0%20%5C%20kg%5C%5C%5C%5CPercent%5C%20yield%3D97.0%5C%25%5C%5C%5C%5CPercent%20%5C%20yield%20%3D%20%5Cfrac%7BActual%20%5C%20yield%7D%7BTheoretical%20%5C%20yield%7D%20%5Ctimes%20100%20%5C%25%20%5C%5C%5C%5CTheoretical%20%5C%20yield%5C%20%20of%5C%20%20Mn%20%3D%20%5Cfrac%7BActual%20%5C%20yield%7D%7BPercent%20%5C%20yield%7D%20%5Ctimes%20100%5C%25%5C%5C%5C%5C)
Equation:
Calculating the amount of
![= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\](https://tex.z-dn.net/?f=%3D%204.12%20%5C%20kg%20%5Ctimes%20%5Cfrac%7B1000%20%5C%20g%7D%7B1%20%5C%20kg%7D%20%5Ctimes%20%5Cfrac%7B1%20%5C%20mol%20%5C%20Mn%7D%7B54.94%20%5C%20g%7D%20%5Ctimes%20%5Cfrac%7B3%20%5C%20Mol%20%5C%20MnO_2%7D%7B3%20%5C%20mol%20%5C%20Mn%7D%20%5Ctimes%20%5Cfrac%7B86.94%20%5C%20g%7D%7B1%20%5C%20mol%20%5C%20MnO_2%20%7D%5C%5C%5C%5C%3D%20%206516%20%5C%20g%20%5C%5C%5C%5C%3D6.52%20%5C%20kg%5C%5C%5C%5C)
Chemical reactions are basically divided into two major classes depending on whether the reaction lose energy or gain energy from the environment during the course of the reaction. The two classes of reaction are exothermic and endothermic reaction.
An exothermic reaction is a type of reaction in which the reaction system lose energy to the environment and thus, the energy content of the reactants is more than that of the product formed. Because of this, the enthapyl change of an exothermic reaction is always negative.
An endothermic reaction is a type of reaction in which the reaction system absorb energy from the environment. Thus, the energy contents of the products is always higher than that of the reactants and the enthapyl change of the reaction is always positive. During the course of the reaction, the reaction container is usually cold to the touch because energy is been absorbed from the environment.
Answer: the answer is c.The unbalanced push causes the cart to speed up.
Explanation:
The application of an unbalanced force (the push) causes the cart to speed up. When the cart is in constant motion, the forces are balanced and there is no speeding up. Once an unbalanced force is added, the cart's speed changes.
Answer:
C = 18.29 g
Explanation:
Given data:
Mass of beryllium needed = ?
Mass of nitrogen = 18.9 g
Solution:
Chemical equation:
3Be + N₂ → Be₃N₂
now we will calculate the number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 18.9 g/ 28 g/mol
Number of moles = 0.675 mol
Now we will compare the moles of nitrogen and Be from balance chemical equation.
N₂ : Be
1 : 3
0.675 : 3/1×0.675 = 2.03 mol
Number of moles of Be needed are 2.03 mol.
Mass of Beryllium:
Mass = number of moles × molar mass
Mass = 2.03 mol × 9.01 g/mol
Mass = 18.29 g