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Licemer1 [7]
3 years ago
11

8.Which arrow or arrows represent a release of carbon dioxide? What process is occurring at the arrow(s) you selected? 9. Which

arrow or arrows indicate a process that cycles carbon from living or nonliving organisms? Describe the process or processes you selected. 10. Which arrow or arrows represent reactions that demonstrate a conservation of mass and energy? Explain your answer.

Chemistry
1 answer:
arsen [322]3 years ago
8 0

Answer:

plant respiration shows arrows where CO2 is released.the arrow is going out of the trees represent the release of carbon dioxide.the arrows from the factories,clouds,water shows the circulation of carbon.

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A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2
a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

<u>Explanation:</u>

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=13.00g

Mass of H_2O=2.662g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 13.00 g of carbon dioxide, \frac{12}{44}\times 13.00=3.54g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 2.662 g of water, \frac{2}{18}\times 2.662=0.296g of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles

  • <u>Step 2:</u> 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 0.295 moles.

For Carbon = \frac{0.295}{0.295}=1

For Hydrogen = \frac{0.296}{0.295}=1

For Oxygen = \frac{0.603}{0.295}=2.044\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Hence, the empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

3 0
4 years ago
How much heat is needed warm 65.34 g of water from 18.43 to 21.75
BaLLatris [955]

Given that  

Mass of water = 65.34 g

Amount of heat = mass of water * specific heat (temperature change

)

= 65.34 g * 4.184 J / g-C ( 21.75-18.43 )C

= 907.63  J  

= 0.908 KJ

And  

1 cal = 4.186798 J

907.63 J * 1 cal / 4.186798 J =216.78  cal

Or0.218 kcal  


5 0
3 years ago
Iodine is a solid nonmetal. What is one property of iodine?
KiRa [710]

Answer:

The correct answer is B. Dull appearance

Explanation:

6 0
3 years ago
Using standard free energy of formation values given in the introduction, calculate the equilibrium constant kp of the reaction
WINSTONCH [101]

d G = 2 (dG NOCl) - 2 (dG NO)

= 2 * 66.08 - 2 * 87.6 = -43.04 kJ /mol

dG = - 2.303 RT log K

-43.04 = - 2.303 RT log K

7.54 = log K

K = 3.42 * 10^7

Kp = k (RT)^dn

d n = 2 - 1 -2 = -1

Kp = (3.42 * 10^7) (8.314 * 298)^-1

Kp = 1.31 * 10^4

6 0
4 years ago
Calculate the frequency in hertz of electromagnetic radiation that has a wavelength of 645.0 nm. (c = 3.00 X 10⁸ m/s)
bazaltina [42]

Answer:

1/wavelength

c=1/alpha

6 0
3 years ago
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