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SpyIntel [72]
3 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]3 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]3 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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Two forces, one of 100 ponds and the other 150 pounds act on the same object, at angles of 20°and 60°, respectively, withthe pos
soldi70 [24.7K]
<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

Explanation:

Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.

That is force 1 is 100 pound with x axis at 20°

           F₁ = 100 cos 20 i  +  100 sin 20 j

           F₁ = 93.97 i  +  34.20 j          

That is force 2 is 150 pound with x axis at 60°

           F₂ = 150 cos 60 i  +  150 sin 60 j

           F₂ = 75 i  +  129.90 j  

F₁ +  F₂ =  93.97 i  +  34.20 j + 75 i  +  129.90 j

F₁ +  F₂ =  168.97 i  +  164.10 j

\texttt{Magnitude = }\sqrt{168.97^2+164.10^2}\\\\\texttt{Magnitude = }235.54pounds\\\\\texttt{Angle = }tan^{-1}\left ( \frac{164.10}{168.97}\right )\\\\\texttt{Angle = }44.16^0

Resultant is 235.54 pounds at an angle 44.16° to X axis.

6 0
3 years ago
What is menat by the age of the universe? how old is the universe?
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The universe is 13.8 billion years old.
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3 years ago
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A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
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kogti [31]

Answer: The temperature after another 5 minutes is 68.5°c

Explanation: Please see the attachments below

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Dinosaurs’ skeletons can be distinguished from those of other reptiles by the structure of the hips and legs.
jeyben [28]
True, they had a hole in their hip socket that allowed them to run faster than other reptiles of their size at the time. As well as most reptiles besides reptiles had legs to the side, rather than under them like dinosaurs did.

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