Answer:
P = 14.1 atm
Explanation:
Given data:
Mass of methane = 64 g
pressure exerted by water vapors = ?
Volume of engine = 24.0 L
Temperature = 515 K
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O + energy
Number of moles of methane:
Number of moles = mass / molar mass
number of moles = 64 g/ 16 g/mol
Number of moles = 4 mol
Now we will compare the moles of water vapors and methane.
CH₄ : H₂O
1 : 2
4 : 2/1×4 = 8 mol
Pressure of water vapors:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
P = 8 mol × 0.0821 atm.L/mol.K× 515 K / 24.0 L
P = 338.25 atm.L/ / 24.0 L
P = 14.1 atm
Explanation:
The study of interaction between the chemical substances is known as chemistry and the person who does this study is called a chemical scientist. Everything around us is made up of matter and the interaction between this matters for one or different substances.
Conduits lead electrical flow in all respects effectively as a result of their free electrons. Separators restrict electrical flow and make poor transmitters. Some basic channels are copper, aluminium, gold, and silver. Some regular encasing are glass, air, plastic, elastic, and wood.
Answer: (Structure attached).
Explanation:
This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.
The aromatic ring in p-fluoroanisole has two sustituents, an <u>halogen</u> and a <u>methoxy group</u>, which are <em>ortho-para</em> directing substituents.
Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using <u>sulfur trioxide as an electrophile</u> (very reactive).
The reaction occurs in three steps:
- The attack on the electrophile forms the sigma complex.
- The loss of a proton regenerates an aromatic ring.
- The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).
Normally, a mixture of <em>ortho-para</em> substituted products would be obtained. However, since both <em>para</em> positions are occupied, only the <em>ortho </em>substituted product is obtained here.
CuCl2 + 2NaNO3 ----> Cu(NO3)2 + 2NaCl
using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g of CuCl2 produces 2(22.98 + 35.45) g of NaCl
134.44 g .................................................... 116.86 g
31.0 g ....................................................31.0 * 116.86 /134.44=26.95g
So percentage yield is 21.2* 100 / 26.95 = 78.7% to nearest tenth
Answer:
11,000 cm
Explanation:
Step 1: Given data
Width of the field (w): 17 meters
Length of the field (l): 38 meters
Step 2: Calculate the perimeter of the field
The field is a rectangle. We can find its perimeter (P) by adding its sides.
P = 2 × w + 2 × l = 2 × 17 m + 2 × 38 m = 110 m
Step 3: Convert the perimeter to centimeters
We will use the relationship 1 m = 100 cm.
110 m × (100 cm/1 m) = 11,000 cm
