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4 years ago

Which explanation is true about what happens to a ray of light when it strikes a mirror?

Chemistry

2 answers:

patriot [66]4 years ago

8 0

For this, the correct option would be (D)

nirvana33 [79]4 years ago

7 0

Answer:

Explanation:

When a rays approaches a plane mirror it is called the incident ray. This incident ray when strikes the surface of the plane mirror gets reflected back and is known as the reflected ray. A line can be drawn perpendicular at he point of incidence on the mirror surface which is called the normal line.

And this normal line perpendicular to the plane mirror divides the angle between the the incident ray and the reflected ray in two equal halves which are known as angle of incidence and angle of reflection respectively.

Now reflection law says that the angle of incidence is equal to the angle of reflection and the ray gets reflected back on opposite to the normal line.

Thus option B is correct.

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The Heliocentric model is when the sun, planets, and stars are orbiting a stationary Earth.

Anarel [89]

False

A heliocentric is when the model is
representing the sun as the center
A geocentric says the earth is at the center of the universe

3 0

3 years ago

On the basis of the Ksp values below, what is the order of the solubility from least soluble to most soluble for these compounds

Viktor [21]

Answer:

The order of solubility is AgBr < Ag₂CO₃ < AgCl

Explanation:

The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:

Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).

It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:

Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.

Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that

Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋

Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12

s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶

And for AgCl

AgCl ⇄ Ag⁺ + Cl⁻

Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵

Therefore, AgCl is more soluble than Ag₂CO₃

The order of solubility is AgBr < Ag₂CO₃ < AgCl

8 0

3 years ago

Read 2 more answers

When active metals such as magnesium are immersed in acid solution, hydrogen gas is evolved. Calculate the volume of H2(g) at 30

V125BC [204]

Answer:

The volume of H₂ (g) obtained is 22.4L

Explanation:

First of all, think the reaction:

2HCl (aq) + Zn (s) → ZnCl₂ (aq) + H₂ (g)

You have to add a 2, in the HCl to get ballanced.

Now we should know how many moles of each reactant, do we have.

Volume . Molarity = moles

Notice that volume is in mL, so I must convert to L.

275 mL = 0.275L

0.275L . 0.725mol/L = 0.2 moles of HCl

Molar mass of Zn: 65.41 g/m

50 g / 65.41 g/m = 0.764 moles

Ratio between reactants is 2:1, so I need the double of moles of HCl to react, and a half moles of Zn to react.

My limiting reactant is the HCl, for 0.764 moles of Zinc, I need 1.528 (0.764 .2) of HCl, and I only have 0.2 moles.

Ratio between HCl and H₂ is 1:1, so 0.2 HCl make 0.2 moles of gas

Now apply the Ideal Gas Law, to find out the volume

P. V = n . R . T

2 atm . V = 0.2 mol . 0.08206L atm/K mol . 273K

V = (0.2 mol . 0.08206L atm/K mol . 273K ) / 2 atm

V = 2.24 L

4 0

3 years ago

Consider the following statistic. As of January 2009, the USA has produced 60,000 metric tons of nuclear waste in 60 years of op

Korolek [52]

Answer: A

Explanation:

4 0

2 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$