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3 years ago

Consider the unbalanced equation for the oxidation of aluminum.

2 answers:

8 0

In order to balance an equation, we apply the principle of conservation of mass, which states that mass can neither be created nor destroyed. Therefore, the mass of an element before and after a reaction remains constant. Here, the balanced equation becomes:
4Al + 3O₂ → 2Al₂O₃

The coefficients are 4, 3 and 2.

Pani-rosa [81]3 years ago

8 0

Answer:

Explanation:

edg 2020

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Describe differences between conduction and convection give examples

nevsk [136]

Explanation:

Both conduction and convection are both forms of heat transfer from one place to another.

In conduction, there must be contact between two bodies for the process to take place but in convection, the matter moves to transfer heat.
Conduction mostly occurs in solid substances whereas convection occurs mostly in fluids.
Heat transfer in conduction is quite slow compared to convection which is much faster.

Example of conduction is heating of iron pot when cooking

Example of convection is the refrigerating system.

4 0

3 years ago

It's 32°F (0°C) outside under normal atmospheric conditions (1 atm) at a stunt

Korolek [52]

The volume of the balloon is approximately 2652 liters.

<h3>How to determine the volume occupied by the gas in a balloon </h3>

Let suppose that <em>flammable</em> hydrogen behaves ideally. GIven the molar mass ( $M$ ), in kilograms per kilomole, and mass of the gas ( $m$ ), in kilograms. The volume occupied by the gas ( $V$ ), in cubic centimeters, is found by the equation of state for <em>ideal</em> gases:

$V = \frac{m\cdot R_{u}\cdot T}{P\cdot M}$ (1)

Where:

$R_{u}$ - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.
$T$ - Temperature, in Kelvin
$P$ - Pressure, in kilopascals

If we know that $m = 0.239\,kg$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}$ , $T = 273.15\,K$ , $P = 101.325\,kPa$ and $M = 2.02\,\frac{kg}{kmol}$ , then the volume of the balloon is:

$V = \frac{(0.239\,kg)\cdot \left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot (273.15\,K)}{\left(101.325\,kPa\right)\cdot \left(2.02\,\frac{kg}{kmol} \right)}$

$V = 2.652\,m^{3}$ ( $2652\,L$ )

The volume of the balloon is approximately 2652 liters. $\blacksquare$

To learn more on ideal gases, we kindly invite to check this verified question: brainly.com/question/8711877

4 0

2 years ago

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$ , $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$ , then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0

2 years ago

For the reaction below, initially the partial pressure of all 3 gases is 1.0atm. . 2NH3(g)--> N2(g) + 3H2(g) K, 0.83 1. When

erma4kov [3.2K]

Answer:

The reaction would shift toward the reactants

When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm

Explanation:

For the reaction:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Where K is defined as:

$K = \frac{P_{N_{2}}*P_{H_2}^3}{P_{NH_3}^2} = 0.83$

As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:

$Q = \frac{1atm*{1atm}^3}{1atm^2} = 1$

As Q > K, <em>the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.</em>

Thus, there are true:

<h3>The reaction would shift toward the reactants</h3><h3>When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm</h3>

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4 0

3 years ago

Describe what happens in a condensation reactions

dlinn [17]

A condensation reaction is described to be a reaction wherein two molecules form an even larger product and consequently produces a smaller molecule as a by-product. For example, when two amino acids are combined, a dipeptide bond is formed. As a result, 1 molecule of water is produced as a by-product.

5 0

3 years ago