Answer: the coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.
Explanation:
Original volume of mercury = 1000 cm3.
The final volume of mercury considering its volume expansion quotient = 1000 + 1000*(1.8*10^-4 *52) = 1000 + 9.36 = 1009.36 cm^3
Considering the glass as a non expanding substance, the complete excess volume of 9.36 cm3 of mercury should have overflown the container, but due to the expansion of glass, the capacity of mercury containment increases and so a lesser amount of mercury flows out.
The amount of mercury that actually flowed out = 8.50 cm3.
So, the expansion of the glass container = 9.36-8.50 = 0.86 cm3.
Using the formula for coefficient of expansion,
coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.
Answer:
Explanation:
We can calculate the volume of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.
We know that volume=4/3×πr³
volume =4/3×π(2.0×10⁻¹⁰m)³
volume=33.40×10⁻³⁰m³
Volume of oxygen molecule=33.40×10⁻³⁰m³
we know the ideal gas equation as:
PV=nRT
k=R/Na
R=k×Na
PV=n×k×Na×T
n×Na=N
PV=Nkt
p is pressure of gas
v is volume of gas
T is temperature of gas
N is numbetr of molecules
Na is avagadros number
k is boltzmann constant =1.38×10⁻²³J/K
R is real gas constant
So to calculate pressure using the formula;
PV=NkT
P=NkT/V
Since there is only one molecule of oxygen so N=1
P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³
p=12.39×10⁷Pascal
Yes. Heating up the solvent gives the molecules more kinetic energy. The more rapid motion means that the solvent molecules collide with the solute with greater frequency and the collisions occur with more force. Both factors increase the rate at which the solute dissolves.
Answer: A. Pollutants
Explanation:
"A pollutant is a substance or energy introduced into the environment that has undesired effects, or adversely affects the usefulness of a resource. ..." - Wikipedia
Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
