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MrRissso [65]
3 years ago
11

Of the three properties of colour, hardness, and reflectivity, which property or properties describe how a material interacts wi

th light? (1 point) SHOW EVIDENCE AND DONT POINT scam. This quiz is a Connexus academy quiz called: Physical and Chemical Properties Quick Check
reflectivity only



hardness and reflectivity only

color and reflectivity only


color, hardness, and reflectivity

Use the following observations about a material to answer the question.

The material is shiny and reflects light.
An iron nail is able to scratch the material.
Which statement best classifies the observations?

(1 point)

Only observation 1 uses a physical property of matter.


Only observation 2 uses a physical property of matter.

Neither observation uses a physical property of matter.


Both observations use a physical property of matter.
Chemistry
1 answer:
hodyreva [135]3 years ago
5 0

The properties that describe how a material interacts with light are color and reflectivity only.

Both observations use a physical property of matter.

The physical properties of substances are the properties of substances that can be observed without changing the arrangement of atoms in the substance.

The ability of an object to scratch another is known as hardness while the ability to reflect visible light is called reflectivity. Both are physical properties.

The color of a substance results from the interaction of the molecules of the substances with light. Objects have specific colors depending on the wavelength of light absorbed and subsequently emitted by the substance.

Reflectivity deals with the ability of an object to reflect visible light.

Both color and reflectivity describe how a material interacts with light.

Learn more; brainly.com/question/81085

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Merry Christmas

6 0
3 years ago
Please help me with this. (: I'd appreciate it.
Soloha48 [4]

Answer:

The answer to your question is:  ΔH = -283 kJ/mol, first option

Explanation:

Reaction

                CO  +  O₂     ⇒     CO₂

ΔH = ∑H products - ∑H products

ΔH = -393.5 - (-110.5 + 0)

ΔH = -393.5 + 110.5

ΔH = -283 kJ/mol

4 0
3 years ago
Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10
kkurt [141]

Answer:

The diameter of the hydrogen \mathbf{d =1.0605 \times 10^{-10}\ m}

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

L = \dfrac{nh}{2 \pi}

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

mvr= \dfrac{nh}{2 \pi}

v= \dfrac{nh}{2 \pi mr} ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}   ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2

ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}

r =\dfrac{n^2h^2}{4 \pi^2 mke^2}

For ground-state n = 1

h = (6.625 \times 10^{-34} \ J.s)^2

m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)

Ke = (1.6 \times 10^{-19} \ C)^2

r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}

r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}

\mathbf{r = 5.3025 \times 10^{-11} \ m}

Therefore, the diameter of hydrogen d = 2r

\mathbf{d = ( 2 \times  5.3025 \times 10^{-11} \ m})}

\mathbf{d =1.0605 \times 10^{-10}\ m}}

4 0
3 years ago
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