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SpyIntel [72]
2 years ago
5

Henry mixed salt and water together in a cup until he observed a clear solution. He measured the mass of the solution. Then he p

laced the cup outside for several sunny days during the summer. After a week, he observed that only solid salt remained in the cup and the mass had decreased. Henry concluded that a physical and chemical change occurred in this investigation.
Which statements correctly defend or dispute his conclusion?

A- He is correct. Dissolving salt in water is a physical change, but evaporating the water is a chemical change. Formation of a solid is evidence that a chemical change occurred.
B- He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.
C- He is incorrect. Dissolving salt in water and evaporation of the water are both physical changes. The reappearance of salt is evidence that the change was reversible by a physical change, so it could not be a chemical change.
D- He is incorrect. Dissolving salt in water and evaporation of the water are both chemical changes. The reappearance of salt is evidence that the change was reversible by a chemical change, so it could not be a physical change.
Physics
2 answers:
Andrei [34K]2 years ago
7 0

Answer:

B) He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.

Explanation:

hope this helps!! :)

Tomtit [17]2 years ago
6 0

Answer:

B- He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.

Explanation: Dissolving Salt in water is a Chemical Change, Because the Salt arrangement is different in solid state than dissolved in water. As we can see in the image below, once the Salt is dissolved, it is separated into its ions, Na+ and Cl- Now, The evaporation process is a physical change, because the water doesn´t change its configuration H20 and it only changes its form, as we can see in the image below.

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GenaCL600 [577]

Answer:

repel each other

Explanation:

The magnitude of the charge of an electron is called... ... If a positively-charged glass rod is suspended so that it turns easily and another positively-charged glass rod is brought close to it, the two rods will... Repel each other.

8 0
2 years ago
A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. When it reaches the maxi
kozerog [31]

Answer:

6.51 m and 37.13 m

Explanation:

from the question we were given

mass of skeet = 0.25 kg

speed of skeet = 25 m/s

angle = 30 degrees to the horizon

mass of pellet = 15 g

speed of pellet = 2000 m/s

without being hit by the pellet, the x and y components of the skeet velocity are  

Vx = 25 cos 30 = 21.65

Vy = 25 sin 30 = 12.5

now

V = U + (a x t)

where V = final velocity, U = initial velocity , a = acceleration, t = time and s = distance

-25 sin 30 = 25 sin 30 + (-9.8 x t)

-12.5 = 12.5 - 9.8 t

t = 2.55 s

also

V^2 = U^2 + 2as  ( s = vertical distance and V = 0 )

0 = (25 sin 30)^2 + 2 x (-9.8) x Y

19.6 Y = 156.25

Y =7.97 m

the distance traveled without the pellet hitting the skeet can be gotten using distance = speed x time

distance = 21.65 x 2.55 = 55.2 m

applying the conservation of linear momentum

on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx)  ...equation 1

on the y axis :   (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2

(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx

 Vx = 20.42m/s

0 + (0.015 x 200) = (0.25 + 0.015) Vy

 Vy = 11.32 m/s

now V^2 = U^2 + 2 as

 0 = 11.3^2 + (2 x (-9.8) x s)

s = 6.51 m                          

  • to find the extra distance moved after collision we apply

s = ut + 1/2 at^2

-7.98 = 11.32t + 1/2 (-9.8) t^2

4.9 t^2 - 11.32t + 7.98

t =  3.17 s

recall that distance = speed x time

distance = 20.42 x 3.17 = 64.73 m

the distance of the skeet before being hit would be half of the distance it travels without being hit, this is because the skeet was hit at its maximum height = 55.2 /2

= 27.6 m

therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m

5 0
3 years ago
Singly charged gas ions are accelerated from rest through a voltage of 10.3 V. At what temperature (in K) will the average kinet
Natasha_Volkova [10]

Answer:

Temperature of the gas molecules is 7.96 x 10⁴ K

Explanation:

Given :

Ions accelerated through voltage, V = 10.3 volts

The work done to change the position of singly charged gas ions is given by the relation :

W = q x V

Here q is charge of the ions and its value is 1.6 x 10⁻¹⁹ C.

Average kinetic energy of gas molecules is given by the relation:

K.E. = \frac{3}{2}kT

Here T is temperature and k is Boltzmann constant and its value is 1.38 x 10⁻²³ J/K.

According to the problem, the average kinetic energy of gas is equal to the work done to move the singly charged ions, i.e. ,

K.E. = W

\frac{3}{2}kT = qV

Rearrange the above equation in terms of T :

T= \frac{2qV}{3k}

Substitute the suitable values in the above equation.

T=\frac{2\times1.6\times10^{-19}\times10.3 }{3\times1.38\times10^{-23} }

T = 7.96 x 10⁴ K

5 0
3 years ago
Can someone help me in this one :)
djyliett [7]
I believe it is “runs on parallel circuits”! my bad if incorrect
3 0
3 years ago
Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with e
antiseptic1488 [7]

Answer:

The length of an edge of each small cube  is 3.43 nm.

Explanation:

Given that,

Temperature of ideal gas =27.0°C

Pressure = 1.00 atm

We need to calculate the length of an edge of each small cube

Using gas equation

PV=nRT

PV=NkT

V=\dfrac{NkT}{P}

For, N = 1

Where,

N = number of molecule

k = Boltzmann constant

T = temperature

P= pressure

Put the value into the formula

V=\dfrac{1\times1.38\times10^{-23}\times(27+273)}{1.03\times10^{5}}

V=4.019\times10^{-26}\ m^3

Now, for the cube

V=L^3

L=V^{\frac{1}{3}}

L=(4.019\times10^{-26})^{\frac{1}{3}}

L=3.43\times10^{-9}\ m

L=3.43 nm

Hence, The length of an edge of each small cube  is 3.43 nm.

3 0
3 years ago
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