la densidad de un aceite liviano para vehiculo es 0.87g/cm3 calcules su volumen especifico y peso especifico

Pe = 8700 N/m³

Vesp = 1,15 cm³/g

Explanation:

Ya que conocemos el valor de la densidad de este aceite, es muy sencillo calcular tanto volumen como peso específico.

En el caso de volumen específico, esta se refiere al cociente del volumen de un líquido con su masa respectiva. En otras palabras, es el inverso de la densidad, por tanto usamos la siguiente expresión para el volumen específico:

Vesp = 1/d

Vesp = 1 / 0,87

Para el caso del peso específico, se calcula como el cociente de la masa del líquido y su volumen que ocupa. Sin embargo no conocemos ninguno de esos dos datos, por lo que para calcular el peso específico solo basta multiplicar este valor por la aceleración de gravedad, que vamos a asumir para este problema que es 10 m/s². Y ademas cambiaremos las unidades de densidad de g/cm³ a kg/m³. Esto es porque las unidades de peso específico son N/m³ y los newton (N) son kg m/s².

d = 0,87 g/cm³ * (1 kg/1000g) * (100 cm/1m)³ = 870 kg/m³

Pe = 870 kg/m³ * 10 m/s²

ESpero te sirva.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

is it a purebred? it looks more like a red nose, but it doesn't have the red could it be a staffordshire bull terrior? unsure if that's the right name for it sorry