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Firlakuza [10]
3 years ago
10

A saturated solution of lead(II) nitrate was made at 50 °C using 100 g of water, and this was then allowed to cool to 30 °C. Pre

dict what would happen as the solution cooled?
Physics
2 answers:
olchik [2.2K]3 years ago
8 0
The solubility of the solution would decrease and some of the lead(II) nitrate would precipitate out 
Sunny_sXe [5.5K]3 years ago
4 0

Answer: The temperature of solution will decease along with the precipitation of solid lead nitrate in solution.

Explanation:

Solubility of the substance is directly effect by the temperature of the solvent.

This is because the kinetic energy of the solvent increase as temperature increase due to which solvent easily breaks apart the solute particles easily and faster which will result in solubility of more solute in solvent and vice versa.

So, when prepared saturated solution of lead nitrate at 50°C was allowed to cool down to 30°C one will observe that the precipitation of solid lead nitrate in solution. And this means that solubility of lead nitrate with decrease in temperature decreases as well.

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5. The wire in consists of two segments of different diameters but made from the same metal. The current in segment 1 is I1. a.
Volgvan

Answer:

hello your question is incomplete attached below is the complete question

answer :

a) I1 = I2

b) J1 > J2

c) E 1 > E2

d) ( vd1 ) > ( vd2 )

Explanation:

a) The currents in the two segments are the same  i.e. I1 = I2  and this is because the segments are connected in series

b) Comparing the current densities J1 and J2 in the two segments

note : current density ∝ 1 / area

The area of the second segment is > the area of first segment  therefore

J1 > J2

J1 ( current density of first segment )

J2 ( current density of second segment )

c) Comparing the electric field strengths E1 and E2

 note : electric field strength ∝ current density

since current density of first segment is > current density of second segment  and conductivity of the materials are the same hence

E 1 > E2

d) Comparing the drift speeds Vd1 and Vd2

( vd1 ) > ( vd2 )

this because  ; vd ∝ current density

7 0
3 years ago
If a thermometer measured the temperature in an oven as 400oF five days in a row when the temperature was actually 397oF, this m
aleksandrvk [35]

Answer:

It can be said to be reliable although it is not valid

Explanation:

This is because Reliability means an indicator of consistency, A measure should produce similar or the same results consistently if it measures the same quantity. So does the thermometer measures over 5days but it is not valid because it deviates from the real value

5 0
3 years ago
What is evaluating?
SSSSS [86.1K]

Answer:

D. Drawing a conclusion about something

3 0
3 years ago
A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
svp [43]

a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

K=\frac{1}{2}mv^2

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






5 0
3 years ago
When a neutrally charged atom loses an electron to another atom, the result is the creation of
asambeis [7]
ionic compound. The atom that lost the electron becomes a cation, and the atom that gains an electron becomes an anion. The cation and anion bond together because they have opposite charges to form an ionic compound. The question may be looking for just cation or anion, though.
8 0
3 years ago
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