How many grams of aluminum oxide would form if 15.2 moles of aluminum burned

Write in exponential form: 34⋅34⋅34⋅34⋅34⋅34⋅34⋅34.

kvv77 [185]

Answer:

34^8

Explanation:

7 0

3 years ago

When a physical change in a sample occurs, which of the following does NOT change?

Vanyuwa [196]

I think thee correct answer from the choices listed above is option D. When a physical change in a sample occurs, composition of the sample does not change. It stays the same. Also, the properties of the sample will still be the same. Hope this answers the question.

8 0

3 years ago

Read 2 more answers

Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

Answer: The standard heat for the given reaction is -138.82 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

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

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ$

Hence, the standard heat for the given reaction is -138.82 kJ

3 0

3 years ago

How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)

zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0

3 years ago

What are three methods of measuring the volume of an object?

Anarel [89]

L×w×h=v
length times width times height equals volume
the order in which you multiply does not matter

5 0

3 years ago