Draw a particulate level drawing of a solid, a liquid, and a gas

Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en

Bezzdna [24]

Answer:

10.19 g CO₂

4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following conversion factors:

1 gallon = 3.785 L

1 L = 1000 mL

Now we convert 1.24 gallons to mL:

1.24 gallon * $\frac{3.785L}{1gallon} *\frac{1000mL}{1L} =$ 4693.4 mL

We calculate the mass of Octane:

4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the stoichiometric ratios and molecular weights to calculate the mass of CO₂ and H₂O:

CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{8molCO_{2}}{1molOctane}$ * 44 g/mol = 10.19 g CO₂

H₂O ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{9molH_{2}O}{1molOctane}$ * 18 g/mol = 4.69 g H₂O

7 0

3 years ago

Select the correct answer from each drop-down menu. At chemical equilibrium, the amount of because .

Elena-2011 [213]

Answer:

The answer that completes the question are in BOLD:

At chemical equilibrium, the amount of PRODUCT AND REACTANT REMAIN CONSTANT because the RATES OF THE FORWARD AND REVERSE REACTIONS ARE EQUAL.

Explanation:

In a reversible chemical reaction, an equilibrium is said to be achieved when the rates of the forward reaction is equal to that of the reverse reaction. A reversible reaction is one in which products are formed from reactants simultaneously with the formation of reactants from products.

The combination of two or more substances called REACTANTS gives rise to another substance called PRODUCT, which can in turn give rise to Reactants again. With time, the rate at which the reactants give rise the products, which is called the FORWARD REACTION will be equal to the rate at which the products give rise to the reactants, which is called REVERSE REACTION. At this point, the chemical reaction is said to be in a STATE OF EQUILIBRIUM.

When the rate at which both reaction occurs becomes equal i.e. at an equilibrium state, the concentration of both the reactants and the products becomes constant i.e. no longer changes. Hence, the amount of the reactants forming the products is the same as the amount of products forming the reactants.

N.B: At chemical equilibrium, the amount of the reactants and products does not necessarily equals zero (0). It simply means that there is no net change in the concentration/amount of both reactants and products.

3 0

3 years ago

Name the following ionic compounds, keeping in mind that a transition metal cation must include its charge. ____________________

Lilit [14]

Answer :

(a) The name for $TiO_2$ is titanium(IV) oxide.

(b) The name for $BaCl_2$ is barium chloride.

(c) The name for $CuCl_3$ is copper(III) chloride.

(d) The name for $KI$ is potassium iodide.

(e) The name for $SrCl_2$ is strontium chloride.

(f) The name for $CuBr_2$ is copper(II) bromide.

Explanation :

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

Ionic compound are usually formed when a metal reacts with a non-metal.

The nomenclature of ionic compounds is given by:

1. Positive ion is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

3. In case of transition metals, the oxidation state are written in roman numerals in bracket in-front of positive ions.

(a) $TiO_2$

$TiO_2$ is an ionic compound because titanium element is a metal and oxygen element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on tin is (+4) and the the charge on oxygen is (-2). Thus, the name for $TiO_2$ is titanium(IV) oxide.

(b) $BaCl_2$

$BaCl_2$ is an ionic compound because barium element is a metal and chlorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on barium is (+2) and the the charge on chlorine is (-1). Thus, the name for $BaCl_2$ is barium chloride.

(c) $CuCl_3$

$CuCl_3$ is an ionic compound because copper element is a metal and chlorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on copper is (+3) and the the charge on chlorine is (-1). Thus, the name for $CuCl_3$ is copper(III) chloride.

(d) $KI$

$KI$ is an ionic compound because potassium element is a metal and iodine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on potassium is (+1) and the the charge on iodine is (-1). Thus, the name for $KI$ is potassium iodide.

(e) $SrCl_2$

$SrCl_2$ is an ionic compound because strontium element is a metal and chlorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on strontium is (+2) and the the charge on chlorine is (-1). Thus, the name for $SrCl_2$ is strontium chloride.

(f) $CuBr_2$

$CuBr_2$ is an ionic compound because copper element is a metal and bromine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on copper is (+2) and the the charge on bromine is (-1). Thus, the name for $CuBr_2$ is copper(II) bromide.

3 0

3 years ago

Which of the following are second-order reactions? a. C2H6 → 2 CH3 b. 2 N2O5 → 4 NO2 + O2 c. 2 HI → H2 + I2 d. 2 N2O → 2 N2 + O2

Klio2033 [76]

Answer:

The answer is b, c, d, e

Explanation:

b. 2 N2O5 → 4 NO2 + O2

r = k [N2O5]^2 --> Second-order regarding global reaction

c. 2 HI → H2 + I2

r = k [HI]^2 --> Second-order regarding global reaction

d. 2 N2O → 2 N2 + O2

r = k [N2O]^2 --> Second-order regarding global reaction

e. 2 NO2 → 2 NO + O2

r = k [NO2]^2 --> Second-order regarding global reaction

8 0

3 years ago

Write and label the density triangle

Slav-nsk [51]

Answer:

It is basically a way of telling you how to solve for different variables in the equation d=m/v

Explanation:

8 0

3 years ago