All categories

3 years ago

Explain how a light ray behaves when it strikes a mirror?

Chemistry

2 answers:

Vinvika [58]3 years ago

5 0

The light ray is reflected when it his a mirror.

Trava [24]3 years ago

5 0

The light ray reflects off the mirror.
The direction of the light ray is determined like this:

pretend there is a perpendicular plane to the mirror at where the light strikes the mirror. the light ray is reflected relative to the perpendicular plane, reflecting it across that line/plane.

You might be interested in

What conclusion can she make from her experiment?

pishuonlain [190]

Answer:

hii samyy

Explanation:

nice killua profile pic :))

5 0

3 years ago

Read 2 more answers

Which statement explains why spontaneous generation is a mistaken idea?

IceJOKER [234]

Your answer is D; Not all living things can maintain homeostasis.

8 0

3 years ago

In writing a chemical equation that produces hydrogen gas, the correct representation of hydrogen gas is

Dennis_Churaev [7]

Answer:

Explanation:

5 0

3 years ago

Which represents the correct equilibrium constant expression for the reaction below? Cu(s)+2Ag+(aq)<->Cu2+(aq)+2Ag(s)

balu736 [363]

K(eq) = concentration of products/concentration of reactant = [Cu+2] / [Ag+]^2
Activity of pure solid and liquid is taken as 1.
Hence last option is correct.
Hope this helps, have a great day ahead!
I hope this helps you alot
:)
:)
:0

4 0

3 years ago

Read 2 more answers

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

3 0

3 years ago