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ziro4ka [17]
3 years ago
5

Describe how light waves interact with the clothes you are wearing and your eyes.

Chemistry
1 answer:
Tom [10]3 years ago
6 0

Answer:

You have read that EM waves can interact with a material medium in

the same ways that mechanical waves do. Three forms of interaction

play an especially important role in how people see light. One form is

reflection. Most things are visible because they reflect light. The two

other forms of interaction are transmission and absorption.

Transmission is the passage of an EM wave

through a medium. If the light reflected from objects did not pass

through the air, windows, or most of the eye, we could not see the

objects. Absorption is the disappearance of an

EM wave into the medium. Absorption affects how things look, because

it limits the light available to be reflected or transmitted.

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What is the density of a gas with a volume of 5.6 FL oz. and a mass of 37 grams?
Assoli18 [71]

Answer:

6.60

Explanation:

4 0
3 years ago
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The geometry of a ______________ makes it a very stable shape.
boyakko [2]

Answer:

triangles

Explanation:

5 0
2 years ago
Roundup, an herbicide manufactured by Monsanto, has the formula C₃H₈NO₅P. How many moles of molecules are there in a 304.3-g sam
jekas [21]
<span>The molar mass is 169.09

304.3g/169.09g = 1.799mol which rounds to 1.800 mol</span>
7 0
3 years ago
How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?
zaharov [31]

Answer:

m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2

Best regards!

7 0
3 years ago
Androstenedione, which contains only carbon, hydrogen, and oxygen, is a steroid hormone produced in the adrenal glands and the g
Gennadij [26K]

Answer: The molecular formula for androstenedione is, C_{19}H_{26}O_2

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=5.527g

Mass of H_2O=1.548g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 5.527 g of carbon dioxide, \frac{12}{44}\times 5.527=1.507g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.548 g of water, \frac{2}{18}\times 1.548=0.172g of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (1.893g)-[(1.507g)+(0.172g)]=0.214g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.507g}{12g/mole}=0.126moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.172g}{1g/mole}=0.172moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.214g}{16g/mole}=0.0133moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0133 moles.

For Carbon = \frac{0.126}{0.0133}=9.5

For Hydrogen  = \frac{0.172}{0.0133}=12.9\approx 13

For Oxygen  = \frac{0.0133}{0.0133}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9.5 : 13 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 19 : 26 : 2

Thus, the empirical formula for the given compound is C_{19}H_{26}O_2

The empirical formula weight of C_{19}H_{26}O_2 = 19(12) + 26(1) + 2(16) = 286 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{286.4}{286}=1

Molecular formula = C_{19}H_{26}O_2

Therefore, the molecular formula for androstenedione is, C_{19}H_{26}O_2

3 0
3 years ago
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