How many moles of oxygen are required in order to produce 4 moles of water?

Alexxx [7]

At room temperature oxygen, $O_2$ is present in the gas phase.

A gas is defined as a substance which is characterized by the rapidly movement of molecules which are widely separated. The shape and volume of gas is not definite that is they take on the shape of the container in which it is placed. The gas molecules have minimum inter-nuclear attractions and thus have high kinetic energy.

Thus, the sentences that describes oxygen at room temperature are:

It has no definite volume and takes the shape of its container.
Its particles move fast enough to overcome the attraction between them.
It has more energy than it would be at a cooler temperature as the kinetic energy is directly proportional to the temperature that is kinetic energy increases on increasing the temperature.

6 0

3 years ago

SO2+02➡SO3 balance chemical equation

Slav-nsk [51]

Answer:

2S02 + O2 ---> 2SO3

Its balanced now as Reactants = Products.

8 0

2 years ago

Dose All plants have a vascular system of pipes that transport fluids]

USPshnik [31]

Answer:

certain types of plants (vascular plants) have a system for transporting water, minerals

4 0

3 years ago

CHEM

castortr0y [4]

Molarity = moles / liter

a) M = 2/4 = 0.5 M

b) Moles = 4/(30 + 16 + 1)
= 0.085
M = 0.085 / 2 = 0.0425 M

c) Moles = 5.85 / (23 + 35.5)
= 0.1
M = 0.1 / 0.4
= 0.25 M

3 0

3 years ago

Read 2 more answers

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago