Answer:
12.5 g of Li are needed in order toproduce 0.60 moles of Li₃N
Explanation:
The reaction is:
6Li(s) + N₂(g) → 2Li₃N(s)
If nitrogen is in excess, the lithium is the limiting reactant.
Ratio is 2:6
2 moles of nitride were produced by 6 moles of Li
Then, 0.6 moles of nitride were produced by (0.6 .6)/ 2 = 1.8 moles of Li
Let's convert the moles to mass → 1.8 mol . 6.94 g/ 1mol = 12.5 g of Li
Answer: The entropy change of the surroundings will be -17.7 J/K mol.
Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol
Amount of Acetone given = 10.8 g
Number of moles is calculated by using the formula:

Molar mass of acetone = 58 g/mol
Number of moles = 
If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then
0.1862 moles will have = 
To calculate the entropy change for the system, we use the formula:

Temperature = 56.2°C = (273 + 56.2)K = 329.2K
Putting values in above equation, we get
(Conversion Factor: 1 kJ = 1000J)
At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,


Answer: 1.2642*10²⁵ on both sides
Explanation:
First check how many moles are there on each side.
Since this is a balanaced equataion the number of moles on each side is the same thus the number of atoms is also same on both sides
There are 3 moles of carbon and 8 moles of hydrogen in C3H8
and 2 moles of oxygen in O2 but there 5 infront so 2*5 is 10
Number of moles on the right is 10+8+3 = 21
Now use Avogrado's constant
21 Moles* (6.02*10²³)/Mol
= 21*6.02*10²³
= 1.2642*10²⁵
0.000001ppm
Explanation:
Mass of fluoride = 500g
Volume of water = 500000liters
Unknown:
Parts per million of fluoride = ?
Solution:
The parts per million is the amount of solute in milligram dissolved in a liter of water or milligram per kilogram of solvent
It is a unit used to express very small concentration.
we need to convert g - mg
500g = 500 x 10⁻³mg = 0.5mg
Concentration in parts per million = 
Concentration in parts per million =
= 0.000001ppm
learn more:
Parts per million brainly.com/question/2854033
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When sodium carbonate is dissolved in water, the equation is
.
When carbon dioxide is placed in water, aqueous carbon dioxide is formed: 
<h3>Dissolution of compounds in water</h3>
Some compounds are water-soluble, some are just partially soluble, while others are insoluble in water. Some soluble or partially soluble substances dissociate in water into their component ions. These substances are said to be ionic.
Sodium carbonate, like every other sodium salt, is soluble in water. It dissolves in water to form an aqueous solution of sodium carbonate.
While in solution, sodium carbonate dissociates into its component ions according to the following equation:

Carbon dioxide, on the other hand, does not dissociate in water. Instead, it dissolves in water where most of it remains as aqueous carbon dioxide in equilibrium with a small amount of hydronium ion and hydrogen carbonate ion.
Since the hydronium and hydrogen carbonate ions formed are so minute, the equation of the reaction can be written as: 
More on the dissolution of substances can be found here: brainly.com/question/28580758
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