What is another name for a homogeneous mixture?

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$

From the reaction, we conclude that

As, 3 mole of $O_2$ react to give 2 mole of $Al_2O_3$

So, 0.188 moles of $O_2$ react to give $\frac{2}{3}\times 0.188=0.125$ moles of $Al_2O_3$

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of $Al_2O_3$ releases amount of heat = 1676 kJ

So, 0.125 mole of $Al_2O_3$ releases amount of heat = $0.125\times 1676kJ=209.5kJ$

Thus, the amount of heat released in the combustion is, 209.5 kJ

4 0

3 years ago

A titration is completed using 10.0 mL of 0.50 M HCl and an unknown concentration of NaOH solution. If the equivalence point is

Drupady [299]

Answer:

0.172 M

Explanation:

7 0

3 years ago

When a solid with a mass of 53.78 g is added to 50.0 ml of water in a graduated cylinder, the volume increases to 56.4 ml. Calcu

Bad White [126]

Answer:

<h3>The answer is 8.40 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

volume = final volume of water - initial volume of water

volume = 56.4 - 50 = 6.4 mL

We have

$density = \frac{53.78}{6.4} \\ = 8.403125$

We have the final answer as

Hope this helps you

3 0

3 years ago

What is [OH-]? Please include work so I know how to answer future questions. Thanks! :)

kramer

Answer:

i know it is alot but that is how are teacher told us to do, hope this is correct

Explanation:

a. 2.6

b. 12.0

Explanation:

a.

First, we will calculate the molar concentration of HCl.

M = mass of HCl / molar mass of HCl × liters of solution

M = 0.40 g / 36.46 g/mol × 4.5 L

M = 2.4 × 10⁻³ M

HCl is a strong monoprotic acid, so [H⁺] = 2.4 × 10⁻³ M. The pH is:

pH = -log [H⁺]

pH = -log 2.4 × 10⁻³ = 2.6

b.

First, we will calculate the molar concentration of NaOH.

M = mass of NaOH / molar mass of NaOH × liters of solution

M = 0.80 g / 40.00 g/mol × 2.0 L

M = 0.010 M

NaOH is a strong base with 1 OH⁻, so [OH⁻] = 0.010 M. The pOH is:

pOH = -log [OH⁻]

pOH = -log 0.010 = 2.0

The pH is:

14.00 = pH + pOH

pH = 14.00 - pOH = 14.00 - 2.0 = 12.0

6 0

3 years ago

1. The

mars1129 [50]

Answer: proton

Explanation:

4 0

4 years ago