Answer:
4.5g/ml
Explanation:
metal sample has a mass = 7.56 g
cylinder previously filled with water = 20.00 mL
final volume in the cylinder = 21.68 mL
density = 4.5g/ml
Going from top to bottom
+ 822 kJ
-2466
+ 411
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HOMEREFERENCEEXAMPLESEXAMPLES OF PHYSICAL PROPERTIES
Examples of Physical Properties
7th grade8th grade9th gradeMiddle SchoolHigh SchoolCollege
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A physical property is any property of matter or energy that can be measured. It is an attribute of matter that can be observed or perceived.
Common Physical Properties
Absorption of electromagnetic - The way a photon’s energy is taken up by matter
Absorption (physical) - Absorption between two forms of matter
Albedo - Reflecting power of a surface
Angular momentum - The amount of rotation of an object
Area - Amount of a two dimensional surface in a plane
Brittleness - Tendency of a material to break under stress
Boiling point - Temperature where a liquid forms vapor
Capacitance - Ability of an object to store an electrical charge
Color - Hue of an object as perceived by humans
Concentration - Amount of one substance in a mixture
Density - Mass per unit volume of a substance
Dielectric constant - Storage and dissipation of electric and magnetic energy
Ductility - Ability of a substance to be stretched into a wire
Distribution - Number of particles per unit volume in single-particle phase space
Efficacy - Capacity to produce an effect
Elasticity - Tendency of a material to return to its former shape
Electric charge - Positive or negative electric charge of matter
Electrical conductivity - A material's ability to conduct electricity
Electrical impedance - Ratio of voltage to AC
Electrical resistivity - How strongly a flow of electric current is opposed
Electric field - Made by electrically charged particles and time-varying magnetic fields.
Electric potential - Potential energy of a charged particle divided by the charge
Emission - Spectrum of frequencies of electromagnetic radiation emitted
Flexibility - Pliability
Flow rate - Amount of fluid which passes through a surface per unit time.
Fluidity - Flows easily
Freezing point - Temperature where a liquid solidifies
Frequency - Number of repetitions in a given time frame
Hardness - How resistant solid matter is to external force
Inductance - When the current changes, the conductor creates voltage
Intrinsic impedance - Ratio of electric and magnetic fields in an electromagnetic wave
Intensity - Power transferred per unit area
Irradiance - Power of electromagnetic radiation per unit area
Length - Longest dimension of an object
Location - Place where something exists
Luminance - Amount of light that passes through a given area
Luminescence - Emission of light not resulting from heat
Luster - The way light interacts with the surface of a crystal, mineral or rock
Malleability - Ability to form a thin sheet by hammering or rolling a material
Magnetic moment - Force that the magnet exerts on electric currents and the torque that a magnetic field exerts on it
Mass - An object's resistance to being accelerated
Melting point - Temperature where a solid changes to a liquid
Momentum - Product of the mass and velocity of an object
Permeability - Ability of a material to support a magnetic field
Smell - Scent or odor of a substance
Solubility - Ability of a substance to dissolve
Specific heat - Heat capacity per unit mass of a material
Temperature - Numerical measure of heat and cold
Thermal conductivity - Property of a material to conduct heat
Velocity - Rate of change in the position of an object
Viscosity - Resistance to deformation by stress
Volume - Space that a substance occupies
Answer:
Average atomic mass = 63.62 amu.
Explanation:
Abundance of Co⁶³ = 69.2%
Abundance of Co⁶⁵ = 30.8%
Atomic mass of Co⁶³ = 63 amu
Atomic mass of Co⁶⁵ = 65 amu
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (63×69.2)+(65×30.8) /100
Average atomic mass = 4359.6 + 2002 / 100
Average atomic mass = 6361.6 / 100
Average atomic mass = 63.62 amu.
Answer:
2,54x10² mmHg
Explanation:
To solve this problem you can use Clausius-Clapeyron equation that serves to estimate vapor pressures or temperatures:
Where:
P1 is 1,00x10² mmHg
ΔHvap is 39,3 kJ/mol
R is gas constant 8,314x10⁻³ kJmol⁻¹K⁻¹
T1 is 34,90°C + 273,15 = 308,05 K
T2 is 54,81°C + 273,15 = 327,96 K
Thus:
Thus, P2 is <em>2,54x10² mmHg</em>
I hope it helps!
