Answer:
Explained below
Explanation:
First of all we need to know that diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration.
Now, the meaning of a substance having higher concentration or lower concentration is that the amount of solute of a given concentration is higher in the higher concentration region than that in the region of lower concentration.
-III I
NH₄⁺
V -II
N₂O₅
I V -II
NaNO₃
-----------------------
-3 +5 +5
Answer:
0.440 moles of NH₃ are produced
Explanation:
First of all, we need to determine the limiting reactant by the stoichiometry.
Equation reaction is: N₂(g) + 3H₂(g) ⟶ 2 NH₃(g)
1 mol of nitrogen needs 3 moles of hydrogen to react
Therefore 0.220 moles of N₂ will need (0.220 . 3) / 1 = 0.660 moles of H₂
As we have 0.717 moles of H₂ and we need 0.660, the hydrogen is the excess reagent, therefore, the N₂ is the limiting reactant
3 moles of H₂ need 1 mol of N₂ to react
Then, 0.717 moles of H₂ will react with (0.717 . 1) / 3 = 0.239 moles of N₂
We do not have enough N₂
After complete reaction → ratio is 1:2
1 mol of N₂ reacts to produce 2 moles of ammonia
Therefore 0.220 moles of N₂ will produce (0.220 . 2) / 1 = 0.440 moles of NH₃
Answer:
The correct answer is 783 kg.
Explanation:
To solve this problem, we first must simply add all three values together. We can perform this operation because all three values have the same units of kilograms.
When we add, we get:
258.3 + 271.23 + 253 = 782.53 kg
However, this answer does not have the correct number of significant digits. When adding and subtracting values, we round our final answer to the least amount of decimal places of the numbers we were adding. In this case, that is 0 decimal places, because 253 doesn’t have a decimal place.
Therefore, we should round 782.53 to the ones place. Since the next number is a 5, we round up.
This gives us 783 kg, which is our final answer.
Hope this helps!
0.01 m 
< 0.03 m 
< 0.04 m urea
As molal concentration rises, so does freezing point depression. It can be expressed mathematically as ΔTf = Kfm.
<h3>What is Colligative Properties ?</h3>
- The concentration of solute particles in a solution, not the composition of the solute, determines a colligative properties .
- Osmotic pressure, boiling point elevation, freezing point depression, and vapor pressure reduction are examples of ligand-like properties.
<h3>What is freezing point depression?</h3>
- When less of another non-volatile material is added, the temperature at which a substance freezes decreases, a process known as Freezing-point depression.
- Examples include combining two solids together, such as contaminants in a finely powdered medicine, salt in water, alcohol in water.
- An significant factor in workplace safety is freezing points.
- If a substance is kept below its freezing point, it may become more or less dangerous.
- The freezing point additionally offers a crucial safety standard for evaluating the impacts of worker exposure to cold conditions.
Learn moree about Colligative Properties here:
brainly.com/question/10323760
#SPJ4
