Answer:
I don't have the number of cubes in each bag, but whichever bag had the most cubes would have the most kinetic energy as it falls
Answer:
6.43 moles of NF₃.
Explanation:
The balanced equation for the reaction is given below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.
Thus, 6.43 moles of NF₃ were obtained from the reaction.
Given:
Mixture of vitamin water = 75% pure water & 25% concentrated vitamin drink.
To find:
Quantity of water to be added to 16 gallons of concentrated vitamin drink to prepare a tank of vitamin water.
Solution:
Let total amount of mixture be x.
25% of x = 16 gallons (from given information)
When 25% of x is 16 gallons, the remaining 75% of x will be calculated as below:
(16/25)*75 = 48
Answer: 75% of x = 48 gallons. This means 48 gallons pure water is required to be added to mixture to prepare a tank of vitamin water.
There are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.
HOW TO CALCULATE MASS:
The mass of a substance can be calculated by multiplying the number of moles in the substance by its molecular mass.
However, given that the number of molecules in calcium hydroxide is 4.07 x 10¹⁵molecules, we need to calculate the number of moles in Ca(OH)2 as follows:
no. of moles of Ca(OH)2 = 4.07 x 10¹⁵ ÷ 6.02 × 10²³
no. of moles = 0.676 × 10-⁸
no. of moles = 6.76 × 10-⁹ moles.
Molar mass of Ca(OH)2 = 74.093 g/mol
Mass of Ca(OH)2 = 74.093 × 6.76 × 10-⁹ moles
Mass = 5.01 × 10-⁷grams.
Therefore, there are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.
Learn more about how to calculate mass at: brainly.com/question/8101390?referrer=searchResults
The reason for adding a limited amount and then an excess amount is that initially a metal hydroxide may form which becomes soluble when more base is added and the metal complex forms.
In qualitative analysis is a common to add the base in drops and then in excess. When added in drops, the metal hydroxide is formed. This metal hydroxide is often insoluble.
After this metal hydroxide is formed, the base could be added in excess such that the metal hydroxide dissolves in the excess base by forming a complex.
For instance;
CuCl2(aq) + 2NaOH(aq) -------> Cu(OH)2(s) + 2NaCl(aq)
Cu(OH)2(s) + 2OH^-(aq) -------> [Cu(OH)4]^2+(aq)
Learn more: brainly.com/question/1527403
