Answer:
R=0.0438 Ω
Explicación:
1) Hallar el área o sección del conductor de cobre, usando esta fórmula:
A=π.r² (Pi x radio al cuadrado)
Debido a que conocemos el diámetro (1.5mm) su radio es la mitad de esto es decir 0.75mm, y lo sustituimos en la fórmula:
A=π.(0.75mm)²
A=π(0.5625mm²)
A=1.7671mm²
2) La resistividad del cobre es: rho = 0,0172 y la incluimos en la fórmula siguiente:
R=p
R=0,0172Ω x
Simplificamos:
R=
El resultado es:
R=0.0438 Ω
Explanation:
First c<span>alculate the mole fraction of each substance:
acetone: 2,88 mol </span>÷ (2,88 mol + 1,45 mol) = 0,665.
cyclohexane: 1,45 ÷ (2,88 mol + 1,45 mol) = 0,335.
Raoult's Law:
P(total) = P(acetone) · χ(acetone) + P(cyclohexane) · χ(cyclohexane).
P(total) = 229,5 torr · 0,665 + 97,6 torr · 0,335.
P(total) = 185,3 torr.
χ for acetone: 229,5 torr · 0,665 ÷ 185,3 torr = 0,823.
χ for cyclohexane: 97,6 torr · 0,335 ÷ 185,3 torr = 0,177.
In general, bonds with an electronegativity difference of 0-0.5 are nonpolar covalent, bonds with an EN difference of 0.5-2.0 are covalent, and anything above 2.0 is considered ionic.
To determine the bond types of the pairs of elements, we will need their EN values. We can subtract their EN values to find their EN difference.
H and Br: 2.96-2.20=0.76
Therefore a bond between H and Br would be moderately polar covalent.
Cl and F: 3.98-3.16=0.82
Therefore this bond is moderately polar covalent.
K and Cl: 3.16-0.82=2.34
Therefore this bond is ionic.
Li and O: 3.44-0.98=2.46
Therefore this bond is ionic.
Br and Br: Because these are the same element, meaning that they have the same EN value, we automatically know that their EN difference would be zero. Therefore, this bond is very covalent.
Answer: The Anode reaction is 
Explanation:
Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.
Oxidation reaction is defined as the reaction in which a substance looses its electrons.Oxidation occurs at anode.
Anode :
Reduction reaction is defined as the reaction in which a substance gains electrons. Reduction occurs at cathode.
Cathode : 
Answer:
1 g
Explanation:
Given data:
Volume of solution = 250 mL (250 mL × 1 L/1000 mL = 0.25 L)
Molarity of solution = 0.1 M
Amount of sodium hydroxide = ?
Solution:
Molarity = number of moles / volume in L
0.1 M = number of moles /0.25 L
Number of moles = 0.1 M × 0.25 L
M = mol/ L
Number of moles = 0.025 mol
Mass of sodium hydroxide:
Mass = number of moles × molar mass
Mass = 0.025 mol ×40 g/mol
Mass = 1 g
