All categories

3 years ago

In response to boyle’s law, the pressure of a gas increases as the volume decreases because

Chemistry

1 answer:

inn [45]3 years ago

8 0

because the gas particles are forced closer together.

You might be interested in

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2 NO ( g

enyata [817]

Answer:

ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)

Explanation:

2 NO (g) + O₂ (g) ⇄ 2NO₂ (g)

Let's apply the thermodynamic formula to calculate the ΔG

ΔG = ΔG° + R .T . lnQ

We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q

How can we know Q? By the partial pressures (Qp)

P NO = 0.450atm

PO₂ = 0.1 atm

PNO₂ = 0.650 atm

Qp = [NO₂]² / [NO]² . [O₂]

Qp = 0.650² / 0.450² . 0.1 = 20.86

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86

ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)

5 0

3 years ago

Pleaseeee helpppppppp

Minchanka [31]

Answer:

C. The third option

Explanation:

7 0

3 years ago

OK! this is for the weebs :D

maw [93]

Answer:

Explanation:3

4 0

2 years ago

Read 2 more answers

If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r

kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)$

Energy absorbed by pork,E = $1.6\times 10^3 kJ$ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

$10\%=\frac{E}{Q}\times 100$

$Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ$

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = $-1.6\times 10^4 kJ$

Moles of propane burnt to produce Q energy =n

$n\times \Delta H_{rxn}=Q$

$n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol$

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

$\frac{1}{3}\times 7.83 mol=23.49 mol$ carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0

2 years ago

What mass of H2O is produced by the combustion of 1.00 mol of CH4?

kkurt [141]

CH4 : H2O
1 : 2

number of moles of H2O = 1.00 x 2
number of moles of H2O = 2.00mol

mass = number of moles x molar mass

mass of H2O = 2.00 x (1 + 1 + 16)
mass of H2O = 36g

8 0

2 years ago