Nitrous oxide .
The Lewis dot structure is attached
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
Water (H
2O) is a polar inorganic compound that is at room temperature a tasteless and odorless liquid, which is nearly colorless apart from an inherent hint of blue. It is by far the most studied chemical compound and is described as the "universal solvent" [18][19] and the "solvent of life".[20] It is the most abundant substance on Earth[21] and the only common substance to exist as a solid, liquid, and gas on Earth's surface.[22] It is also the third most abundant molecule in the universe.[21]
Water (H
2O)


NamesIUPAC name
water, oxidane
Other names
Hydrogen hydroxide (HH or HOH), hydrogen oxide, dihydrogen monoxide (DHMO) (systematic name[1]), hydrogen monoxide, dihydrogen oxide, hydric acid, hydrohydroxic acid, hydroxic acid, hydrol,[2] μ-oxido dihydrogen
Identifiers
CAS Number
7732-18-5 
3D model (JSmol)
Interactive image
Beilstein Reference
3587155ChEBI
CHEBI:15377 
ChEMBL
ChEMBL1098659 
ChemSpider
937 
Gmelin Reference
117
PubChem CID
962
RTECS numberZC0110000UNII
059QF0KO0R 
InChI
InChI=1S/H2O/h1H2 
Key: XLYOFNOQVPJJNP-UHFFFAOYSA-N 
SMILES
O
Properties
Chemical formula
H
2OMolar mass18.01528(33) g/molAppearanceWhite crystalline solid, almost colorless liquid with a hint of blue, colorless gas[3]OdorNoneDensityLiquid:[4]
0.9998396 g/mL at 0 °C
0.9970474 g/mL at 25 °C
0.961893 g/mL at 95 °C
Solid:[5]
0.9167 g/ml at 0 °CMelting point0.00 °C (32.00 °F; 273.15 K) [a]Boiling point99.98 °C (211.96 °F; 373.13 K) [6][a]SolubilityPoorly soluble in haloalkanes, aliphaticand aromatic hydrocarbons, ethers.[7]Improved solubility in carboxylates, alcohols, ketones, amines. Miscible with methanol, ethanol, propanol, isopropanol, acetone, glycerol, 1,4-dioxane, tetrahydrofuran, sulfolane, acetaldehyde, dimethylformamide, dimethoxyethane, dimethyl sulfoxide, acetonitrile. Partially miscible with Diethyl ether, Methyl Ethyl Ketone, Dichloromethane, Ethyl Acetate, Bromine.Vapor pressure3.1690 kilopascals or 0.031276 atm[8]Acidity (pKa)13.995[9][10][b]Basicity (pKb)13.995Conjugate acidHydroniumConjugate baseHydroxideThermal conductivity0.6065 W/(m·K)[13]
Refractive index (nD)
1.3330 (20 °C)[14]Viscosity0.890 cP[15]Structure
Crystal structure
Hexagonal
Point group
C2v
Molecular shape
Bent
Dipole moment
1.8546 D[16]Thermochemistry
Heat capacity (C)
75.375 ± 0.05 J/(mol·K)[17]
Std molar
entropy (So298)
69.95 ± 0.03 J/(mol·K)[17]
Std enthalpy of
formation (ΔfHo298)
−285.83 ± 0.04 kJ/mol[7][17]
Gibbs free energy (ΔfG˚)
−237.24 kJ/mol[7]
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6
