Which statement is true?

Use the mass spectrum of europium to determine the atomic mass of europium. where the peak representing eu-151 has an exact mass

slavikrds [6]

Answer: The atomic mass of a Europium atom is 151.96445 amu.

From the given information:

Percent intensity is 91.61% of Europium atom of molecular weight 150.91986 amu.

Percent intensity is 100.00% of Europium atom of molecular weight 152.92138 amu.

Abundance of Eu-151 atom:

$X_{Eu-151}=\frac{0.9161}{0.9161+1.000}=0.4781$

Abundance of Eu-153 atom:

$X_{Eu-153}=\frac{1.000}{0.9161+1.000}=0.5219$

Atomic mass of Europium atom:

$A=(X_{Eu-151}\times150.91986+X_{Eu-153}\times152.92138)amu\\A=(0.4781\times150.91986+0.5219\times152.92138)amu=151.96445 amu$

Therefore, the atomic mass of a Europium atom is 151.96445 amu.

3 0

3 years ago

Read 2 more answers

You need a 450 microgram sample of gold, but you only have a mass balance that measures in decigrams. Convert the amount of gold

sveta [45]

The amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.

<h3>How to convert micrograms to decigrams?</h3>

According to this question, 450 micrograms of a sample of gold is needed but we only have a mass balance that measures in decigrams.

This means that we are to convert the amount of gold you need to decigrams by comparing the exponents.

The conversion factor of micrograms to decigrams is as follows:

1 micrograms = 1 × 10-⁵ decigrams

This means 450 micrograms is equivalent to 450 × 1 × 10-⁵ = 4.5 × 10-³ decigrams

Therefore, the amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.

Learn more about decigrams at: brainly.com/question/6869599

#SPJ1

7 0

1 year ago

La tension que se transmite en la cuerda BD es de 75 lb. Calcula el momento de fuerza generada por la cuerda respecto al punto C

Bas_tet [7]

Answer:

Mc = 1920[lb*in]

Explanation:

Para poder solucionar este problema debemos realizar un análisis estático, por tal motivo lo primero es realizar un diagrama de cuerpo libre con las respectivas fuerzas actuando sobre la barra ABC. DE igual manera calcular la geometría de la configuración mostrada.

El diagrama de cuerpo libre se puede ver en la imagen adjunta, con la solución de este problema.

Lo primero es determinar el angulo t, el cual por medio de las propiedades del triangulo rectángulo se puede determinar.

Con este angulo (t) ya determinado, fijamos la atención en el triangulo BCD, este triangulo no es rectángulo, pero por medio de la ley de senos podemos determinar el angulo omega.

Después de determinar el angulo omega, restamos el angulo (t) para poder determinar el angulo (a).

Seguidamente realizamos una sumatoria de momentos alrededor del punto C, utilizado las respectivas fuerzas con los ángulos descompuestos.

El momento en el punto C es de 1920 [Lb*in].

Nota: ya que no se menciona la fuerza en el punto A, esta se desprecia y no se tiene en cuenta en los calculos. En la imagen adjunta se puede ver el procedimiento desarrollado.

7 0

3 years ago

on aircraft carriers, catapults are used to accelerate jet air craft to flight speeds in a short distance. One such catapult tak

sineoko [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

Acceleration = (70 / 2.5) m/s²

<em>Acceleration = 28 m/s²</em>

That's about 2.9 G's . Jet pilots can endure a lot more than that, but maybe the catapult or the hook on the airplane can't. Let's look a little closer:

F = m A (Newton #2)

The force on the airplane = (18,000 kg) x (28 m/s²)

Force on the airplane = 504,000 Newtons

That's about 113,000 pounds ! Maybe the part of the airplane that the catapult pushes on can't handle any more force than that. Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds. Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes ! Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

8 0

3 years ago

A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with

Dvinal [7]

Answer:

8 seconds

Explanation:

Answer:

Explanation:

Going up

Time taken to reach maximum height= usin∅/g

=3 secs

Maximum height= H+[(usin∅)²/2g]

=80+[(60sin30)²/20]

=125 meters

Coming Down

Maximum height= ½gt²

125= ½(10)(t²)

t=5 secs

6 0

4 years ago