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What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o

Neporo4naja [7]

Answer:

49671 L is the produced volume of ammonia

Explanation:

We think the reaction of ammonia 's production:

N₂(g) + 3H₂(g) → 2NH₃ (g)

We have the moles of each reactant so let's determine the limiting reactant:

Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂

Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂

We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

1 mol of N₂ can produce 2 moles of ammonia

Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃

We apply now, the Ideal Gases Law → P . V = n . R .T

V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

V = 49671 L

We confirm that the nitrogen was the limiting reactant

3 moles of H₂ need 1 mol of nitrogen to react

Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

6 0

3 years ago

Classify each of the following particulate level illustrations as a representation of either a pure substance, a homogeneous mix

Anarel [89]

Answer:

The classification and illustrations are attached in the drawing.

Explanation:

It is possible to identify the pure substance observing the figure, since it is the only one that has 2 joined atoms (purple and blue) which forms a single compound.

On the other hand, the homogeneous mixture is identified by noting that its atoms are more united with respect to the heterogeneous mixture, highlighting that in homogenous mixtures the atoms, elements or substances are not visible to the naked eye and are in a single phase, instead in the heterogeneous mixture if they can be differentiated.

4 0

3 years ago

The vapor pressure of water at 25.0°c is 23.8 torr. determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.

svp [43]

Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958 = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass
= 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water
= 4.7 / 0.88 = 5.34 moles
∴ moles of the solution = total moles in solu - moles of water
= 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
= 53.8 / 0.64 = 84 g/mole

Q: C

moles of urea (NH2)2 CO = mass weight / molar mass
= 4.49 g / 60 g /mol
= 0.07 mol
moles of methanol = mass weight / molar mass
= 39.9 g / 32 g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg
∴Pv(solu) = 84.55 mmHg

7 0

3 years ago

A schoolbus accelerates to 65 mph and enters the freeway. It travels

Inessa [10]

Answer:

Distance cover by school bus = 149.5 miles

Explanation:

Given:

Velocity of school bus = 65 mph

Time taken by school bus = 2.3 hours

Find:

Distance cover by school bus

Computation:

Distance cover = Velocity x Time taken

Distance cover by school bus = Velocity of school bus x Time taken by school bus

Distance cover by school bus = 65 x 2.3

Distance cover by school bus = 149.5 miles

7 0

3 years ago

Classify each process as an endothermic or exothermic process. drag the appropriate items to their respective bins. helpreset ex

myrzilka [38]

Answers:

1) Breaking Solvent-Solvent Attractions is an Endothermic Process.

2) Breaking Solute-Solute Attractions is an Endothermic Process.

3) Forming Solute-Solvent Attractions is an Exothermic Process.

Explanation:
When a solute is dissolved in solvent it either releases heat or absorbs heat depending upon the the interactions broken and interactions formed. At first, the solvent solvent interactions are broken , this process requires heat which is provided either from external source or is provided by the forming of solute solvent bond forming process which is exothermic.

When the solvent molecules get apart the solute particles enter to form interactions with elimination of heat. So, if the heat required to break solvent solvent interactions is greater than the heat provided by solute solvent interactions formation then the solute will not dissolve at room temperature and vice versa.

7 0

3 years ago

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