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const2013 [10]
3 years ago
13

Anna studies the color of her bedroom wall using different lights. In sunlight, she sees that the wall is white. Anna darkens th

e room and shines a beam of light that includes all the wavelengths of visible light onto the wall. In this light, what’s the wall color that Anna sees?
no color

all seven colors of visible light

red

white

black
Chemistry
2 answers:
erica [24]3 years ago
6 0

Answer:

white

Explanation:

Marizza181 [45]3 years ago
4 0

Answer: C white

Explanation:

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Consider the reaction: 2 SO2(g) + O2(g) <----> 2 SO3. If, at equilibrium at a certain temperature, [SO2] = 1.50 M, [O2] =
vlada-n [284]

Answer:

5.79

Explanation:

The following data were obtained from the question:

Concentration of SO2, [SO2] = 1.50 M

Concentration of O2, [O2] = 0.120 M

Concentration of SO3, [SO3] = 1.25 M

Equilibrium constant, (K) =..?

The balanced equation for the reaction is given below:

2SO2(g) + O2(g) <==> 2SO3(g)

From the balanced equation, the equilibrium constant K is written as follow:

K = [SO3]²/ [SO2]²• [O2]

With the above, the value of the equilibrium constant, K can be obtained as follow

K = [SO3]²/ [SO2]²• [O2]

K = (1.25)² / (1.50)² × 0.120

K = 1.5625 / ( 2.25 x 0.120)

K = 5.79

Therefore, the equilibrium constant is 5.79.

7 0
3 years ago
Pls help this my final 20 points!!!! Be right
Anastaziya [24]

Answer:

I think it's B

Explanation:

I hope it helps

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2 years ago
Convert 5.0 x 1027 r.p. of copper to moles of copper.
PtichkaEL [24]
Convert 5.0 to 1027 an you should get your answer
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What is the Mass of a Penny
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2.5 g is the mass of the penny for Canadian coins which have become defunct

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Read 2 more answers
15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula
otez555 [7]

Answer:

OsCO or COOs

Explanation:

Data given

Carbon = 15.89 %

Oxygen = 21.18 %

Osmium = 62.93%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

15.89 g of C     = 15.89 % Carbon

21.18 g of O      =   21.18 % Oxygen

62.93 g of Os  =   62.93% Osmium

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  15.89 g/ 12 g/mol

                  no. of mole =  1.3242

mole of C = 1.3242

For Oxygen

Molar mass of O = 16 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  21.18 g/ 16 g/mol

                  no. of mole =  

mole of O = 1.3238

For Os

Molar mass of Os = 190 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  62.93 g/ 190 g/mol

                  no. of mole =  

mole of Os = 1.3312

Now we have values in moles as below

C = 1.3242

O = 1.3238

Os = 1.3312

Divide the all values on the smalest values to get whole number ratio

C = 1.3242 /1.3238 = 1.0003

O = 1.3238 /1.3238 = 1

Os = 1.3312 /1.3238 = 1.0056

So all have round value 1 mole

C = 1

O = 1

Os = 1

So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio

7 0
3 years ago
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