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1 year ago

The group of molecules called nucleotides contain:

1 answer:

8 0

The group of molecules called nucleotides contain phosphate groups, pyrimidines, purines, and pentose (a 5-carbon sugar). Therefore, E. all of the above is the correct answer.

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Calculate the number of grams of carbon dioxide produced from complete combustion of one liter of octane by placing the conversi

lesya [120]

Answer:

15.71g

Explanation:

The general combustion equation for all hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

For octane, C8H18 :

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (balanced)

From the balanced equation,

2 x 22.4 L of octane produced 16 [ 12 + (16 x 2)] of carbon dioxide

That is,

44.8 L of octane produced 704g of carbon dioxide

So, 1L of octane will produce 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Therefore, 15.71g of carbon dioxide will be produced by the complete combustion of 1 L of octane.

7 0

2 years ago

How do combination circuits help prevent problems in circuits in a home? If it helps, think about a bedroom since all electrical

den301095 [7]

‍♀️ ◻️✅◾️▫️‍◾️◻️◻️◼️

6 0

1 year ago

A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to

melomori [17]

Answer:The change in temperature of the gas mixture is -68.44 K.

Explanation:

1. Temperature of the bottle when only helium gas was present :T

Pressure inside the container =P= 1.83 atm

Volume of the container = V = 2.50 L

Moles of helium gas = n = 0.458 moles

Using an Ideal gas equation:

$PV=nRT$

$T=\frac{PV}{nR}=\frac{1.83 atm\times 2.50 L}{0.458 mol\times 0.0820 atm L/mol K}=121.81 K$

2. Temperature of the bottle when argon gas is added into the container:T'

Pressure inside the container = P' =2.05 atm

Volume of the container = V = 2.50 L

Moles of helium gas = n' = 0.458 mol + 0.713 mol = 1.171 mol

$P'V=n'RT'$

$T'=\frac{P'V}{n'R}=\frac{2.05 atm\times 2.50 L}{1.171 mol\times 0.0820 atm L/mol K}=53.37 K$

The change in temperature will be given as:

Final temperature = Initial temperature:T' - T

53.37 k - 121.81 K = -68.44 K

The change in temperature of the gas mixture is -68.44 K.

7 0

2 years ago

Read 2 more answers

Given that 2S (s)+3O2 (g)→2SO3 (g)2SO2 (g)+O2 (g)→2SO3 (g) has an enthalpy change of −790.4 kJ has an enthalpy change of −198.2

abruzzese [7]

Answer:

The heat of formation of SO2 is -296.1 kJ

Explanation:

Step 1: Data given

2S (s)+3O2 (g)→2SO3 (g) ΔH = -790.4 kJ

2SO2 (g)+O2 (g)→2SO3 (g) ΔH = -198.2 kJ

Step 2: Calculate the heat of formation of SO2

2 S(s) + 3 O2(g) --> 2 SO3(g) ΔH = -790.4 kJ

S(s) + 3/2 O2(g) → SO3(g) ΔH = -395.2 kJ

2SO2 (g)+O2 (g)→2SO3 (g) ΔH = -198.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g) ΔH = 99.1 kJ

----------------------------------------------------------------

S(s) + 3/2 O2(g) → SO3(g) ΔH = -395.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g) ΔH = 99.1 kJ

-------------------------------------------------------------------

S (s)+O2 (g)→SO2 (g)

ΔHrxn = (-790.4 /2) kJ + (198.2/2) kJ

ΔHrxn = -395.2 kJ + 99.1 kJ = 296.1 kJ

The heat of formation of SO2 is -296.1 kJ

4 0

2 years ago

Movement of the ___ creates the London dispersion forces.

Tanzania [10]

Answer: electrons

Explanation: moving electrons cause momentarily charge

Distribution on molecule. This distribution induces similar distribution to

Adjacent molecule.

7 0

2 years ago