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4 years ago

Give the symbols for 4 species that are isoelectronic with the telluride ion, te2-.

1 answer:

3 0

<span>Answer: Chlorine has 17 electrons therefore for 1+ and 2+ we need the atoms that have 18 and 19 electrons these are argon and potassium: Ar+ and K 2+ For 1- and 2- we need the atoms that have 16 and 15 electrons and these are sulfur and phosphorus, S- and P 2-. Note that + ions imply the loss of electrons and - ions equate to the gain of electrons.</span>

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A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

3 0

3 years ago

A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per

FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0

3 years ago

A 4kg block sitting on the floor, how much potential energy does it have?

prohojiy [21]

Well, there you have a very important principle wrapped up in that question.

There's actually no such thing as a real, actual amount of potential energy.
There's only potential <em><u>relative to some place</u></em>. It's the work you have to do
to lift the object from that reference place to wherever it is now. It's also
the kinetic energy the object would have if it fell down to the reference place
from where it is now.

Here's the formula for potential energy: PE = (mass) x (gravity) x (<em><u>height</u></em><u>)</u> .

So naturally, when you use that formula, you need to decide "height above what ?"

If you're reading a book while you're flying in a passenger jet, the book's PE is
(M x G x 0 meters) relative to your lap, (M x G x 1 meter) relative to the floor of the
plane, (M x G x 10,000 meters) relative to the ground, and maybe (M x G x 25,000 meters)
relative to the bottom of the ocean.

Let's say that gravity is 9.8 m/s² .

Then a 4kg block sitting on the floor has (39.2 x 0 meters) PE relative to the floor
it's sitting on, also (39.2 x 3 meters) relative to the floor that's one floor downstairs,
also (39.2 x 30 meters) relative to 10 floors downstairs, and if it's on the top floor of
the Amoco/Aon Center in Chicago, maybe (39.2 x 345 meters) relative to the floor
in the coffee shop that's off the lobby on the ground floor.

3 0

3 years ago

Read 2 more answers

A 0.0575 kg ice cube at −30.0°C is placed in 0.557 kg of 35.0°C water in a very well insulated container, like the kind we used

Elan Coil [88]

Answer:

t= 22.9ºC

Explanation:

Assuming no heat exchange outside the container, before reaching to a condition of thermal equilibrium, defined by a common final temperature, the body at a higher temperature (water at 35ºC) must give heat to the body at a lower temperature (the ice), as follows:

Qw = c*m*Δt = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t) (1)

This heat must be the same gained by the ice, which must traverse three phases before arriving at a final common temperature t:

1) The heat needed to reach in solid state to 0º, as ice:

Qi =ci*m*(0ºC-(-30ºC) = 0.0575kg*2090(J/kg.ºC)*30ºC = 3605.25 J

2) The heat needed to melt all the ice, at 0ºC:

Qf = cfw*m = 3.33*10⁵ J/kg*0.0575 kg = 19147.5 J

3) Finally, the heat gained by the mass of ice (in liquid state) in order to climb from 0º to a final common temperature t:

Qiw = c*m*Δt = 4190 (J/kg.ºC)*0.0575 kg*(t-0ºC)

So, the total heat gained by the ice is as follows:

Qti = Qi + Qf + Qiw

⇒Qti = 3605.25 J + 19147.5 J + 240.9*t = 22753 J + 240.9*t (2)

As (1) and (2) must be equal each other, we have:

22753 J + 240.9*t = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t)

⇒ 22753 J + 240.9*t = 81684 J -2334*t

⇒ 2575*t = 81684 J- 22753 J = 58931 J

⇒ $t= \frac{58931J}{2575 J/C} = 22.9C$

⇒ t = 22.9º C

3 0

3 years ago

A piece of thin uniform wire of mass m and length 3b is bent into an equilaeral triangle.

Nookie1986 [14]

Answer: Option (C) is the correct answer.

Explanation:

Formula for moment of inertia is as follows.

M.I = $\frac{mass \times (length)^{2}}{3}$

Hence, moment of inertia of two rods is as follows.

M.I of two rods = $2 \times \frac{(\frac{m}{3} \times b^{2})}{3}$

= $\frac{2mb^{2}}{9}$

As third rod have no connection with vertices. So, moment of inertia of a rod along an axis passing through its center is as follows.

M.I = $\frac{mass \times (length)^{2}}{12}$

= $\frac{mb^{2}}{3 \times 12}$

= $\frac{mb^{2}}{36}$

Using parallel axis theorem moment of inertia through vertices is as follows.

$\frac{mb^{2}}{36} + mass \times \text{distance between the two axes}$

$h^{2} = b^{2} - \frac{b^{2}}{4}$

= $\frac{3b^{2}}{4}$

h = $\frac{\sqrt{3b}}{2}$

Now, we will calculate the moment of inertia of third rod about vertices is as follows.

$\frac{mb^{2}}{36} + [(\frac{m}{3}) \times 3\frac{b^{2}}{4}]$

= $mb^{2}[\frac{1}{36} + \frac{1}{4}]$

= 5 $\frac{mb^{2}}{18}$

Therefore, total moment of inertia is calculated as follows.

Total M.I = $\frac{2mb^{2}}{9} + \frac{5mb^{2}}{18}$

= $\frac{mb^{2}}{2}$

Thus, we can conclude that the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices is $\frac{mb^{2}}{2}$ .

4 0

3 years ago