All categories

3 years ago

Describe a homogeneous mixture

Chemistry

2 answers:

hoa [83]3 years ago

8 0

Its when to chemicals are the same and can be taken out of but the product is still the same

ANTONII [103]3 years ago

6 0

It's a mixture that is the same through out. NO clumps or layers. It can have multiple elements/substances combined within.

Hope this helps!

You might be interested in

(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed

soldier1979 [14.2K]

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 KClO₃ → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

KClO₃: 2 moles
KCl: 2 moles
O₂: 3 moles

The molar mass of the compounds is:

KClO₃: 122.45 g/mole
KCl: 74.45 g/mole
O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
KCl: 2 moles ×74.45 g/mole= 148.9 grams
O₂: 3 moles ×32 g/mole= 96 grams

<h3>Ideal gas law</h3>

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that the gas occupies.
T is the temperature of the gas.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

<h3>Number of O₂ produced.</h3>

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
V= 165 mL= 0.165 L (being 1000 mL= 1 L)
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 $\frac{atmL}{molK}$ × 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 $\frac{atmL}{molK}$ × 303 K)

<u><em>n= 0.0059 moles</em></u>

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

<h3>Mass of KClO₃ required</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

$mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}$

<u><em>mass of KClO₃= 0.48144 grams</em></u>

Finally, 0.48144 grams of KClO₃ was decomposed.

Learn more about

the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

4 0

2 years ago

The enthalpy of combustion of naphthalene (MW = 128.17 g/mol) is -5139.6 kJ/mol. How much energy is produced by burning 0.8210 g

bazaltina [42]

The energy produced by burning : -32.92 kJ

<h3>Further explanation</h3>

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The enthalpy and heat(energy) can be formulated :

$\tt \Delta H=\dfrac{Q}{n}\rightarrow n=mol$

The enthalpy of combustion of naphthalene (MW = 128.17 g/mol) is -5139.6 kJ/mol.

The energy released for 0.8210 g of naphthalene :

$\tt Q=\Delta H\times n\\\\Q=-5139.6~kJ/mol\times \dfrac{0.8210~g}{128.17~g/mol}~\\\\Q=-32.92~kJ$

3 0

3 years ago

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:

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