Octane (C8H18) is a component of gasoline that burns

What reaction is depicted by the given equation: au3 3e− au

mrs_skeptik [129]

This is a reduction half-reaction. Initially, gold exists as $Au^{3+}$ , and after the reaction, it has been converted to elemental gold. To make this possible, three electrons must be added to $Au^{3+}$ . This fits within the definition of reduction reactions, where the reactant gains electrons in the reaction.

3 0

3 years ago

Balancing Al+ Agcl, i have problems with this so help!!!

EleoNora [17]

Answer:

Al + 3AgCl → AlCl₃ + 3Ag

Explanation:

The given equation is:

Al + AgCl →

We are to find the product and hence balance the equation. This problem is a simple single replacement reaction.

By virtue of this, Aluminum will displace Ag from the solution:

Al + AgCl → AlCl₃ + Ag

We then balance the equation:

Al + 3AgCl → AlCl₃ + 3Ag

5 0

3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago

Given the following chemical reaction: 2 O2 + CH4 --> CO2 + 2 H2O

Elis [28]

The answer is D) 144 grams O2

5 0

3 years ago

Read 2 more answers

Please help me. Thank you so much to whoever decides to help me and I love you

Bumek [7]

Answer:

2,702.7027 I think ...

3 0

3 years ago