PLEASE HELP ME ASAP!!! I need the answer to #9. If you can show work please do.

How many moles of O2 should be supplied to burn 1.82 mol of C3H8 (propane) molecules In a camping stove?

pashok25 [27]

C3H8 + 5O2 ------> 3CO2 + 4H2O

from reaction 1 mol 5 mol

given 1.82 mol x mol

x=(1.82*5)/1 = 9.10 mol CO2

4 0

3 years ago

Which of the following includes the correct pairing of volcano and its formation? Cinder volcano; large, explosive eruptions Cin

Anarel [89]

The correct pairing of the volcano to its formation would be the composite volcano wherein it usually yields large and violent eruptions. In addition to that, composite volcanoes or also known as the stratovolcano is usually made of materials containing increasing layers of hardened lava.

4 0

3 years ago

Read 2 more answers

Suppose that you want to compare the mass of a block of ice to its mass as liquid

Morgarella [4.7K]

Answer:

You will find the mass of the pan and water but if the water got to its boiling temperature that mass may be a little bit off seeing as some of it may have evaporated

5 0

3 years ago

There are 55.0 g of neon gas in a 3.0 L cylinder, the pressure is 4.5 atm what is the temp of the gas

marshall27 [118]

Answer:

60.9 Kelvin

Explanation: First, write out everything that you know. You are tring to find the temperature, so the temperature will be represented by x.

Pressure (P)= 4.5 atm

Volume (V)= 3L

Number of Moles (n)= ?

Gas Consant (R)= 0.0821, if the pressure is in atm, that means r is 0.0821

Temperature (T)= x

We don't have all the information we need to plug the values into the equation. We still need to know how many moles 55.0 grans of neon is.

Ne in Grams= 55

Atomic Mass of Ne= 20.1797

55/20.1797= 2.7

moles= 2.7

Now that we have all the information we need, plug everying into the equation. In case you don't know, the Ideal Gas Law Equation is PV= nRT.

(4.5)(3) = (2.7)(0.821)x

x= 60.9

Now you have your temperature! It is 60.9 in Kelvin.

4 0

3 years ago

1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change

Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

1) To show that heat flows spontaneously from high temperature to low temperature

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) Entropy change for Decomposition of mercuric oxide

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

8 0

3 years ago