Which of the 3 molecules: Methane, Propane, Butane would release the most energy when combusted?

How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?

solong [7]

Answer:

Mass= 2.77g

Explanation:

Applying

P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28

PV=nRT

NB

Moles(n) = m/M

PV=m/M×RT

m= PVM/RT

Substitute and Simplify

m= (2.09×1.13×28)/(0.082×291)

m= 2.77g

6 0

3 years ago

I NEED HELP PLEASE, THANKS!

ira [324]

Answer:

About 0.652

Explanation:

Because the reaction is balanced, we can go straight to the next step. The molar mass of potassium is about 39.098, while the molar mass of hydrogen gas is 2 and the molar mass of water is 18. Therefore, 25.5g of potassium would be about 0.652 moles, and 220 grams of water would be about 12.222 moles, making potassium the limiting reactant. Since there is a single unit of each compound on both sides of the equation, there would be an equal amount of moles of potassium and hydrogen, and therefore about 0.652 moles of hydrogen gas would be produced. Hope this helps!

6 0

2 years ago

Name the following molecule in the image

Vesna [10]

$\\ \sf{:}\longrightarrow 2,4-diethyl-2-methylpentane$

4 0

2 years ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

V125BC [204]

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 $\frac{cal}{gC}$ .

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

Q is the heat exchanged by a body of mass m.
C is the specific heat substance.
ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

Mass of metal = 50 g
Initial temperature of metal= 45 °C
Final temperature of metal= 11.08 ºC
Specific heat of metal= ?

For water:

Mass of water = 250 g
Initial temperature of water= 10 ºC
Final temperature of water= 11.08 ºC
Specific heat of water = 1.035 $\frac{cal}{gC}$

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 $\frac{cal}{gC}$ × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= $\frac{279.45 cal}{1696 gC}$