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3 years ago

A car with a 14 gallon gas tank gets 22 miles per gallon. How far can the car travel on a full tank of gas?

Chemistry

1 answer:

Sidana [21]3 years ago

3 0

Answer:

308miles

Explanation:

Given parameters:

Volume of tank = 14gallon

Rate of fuel usage = 22miles per gallon

Unknown:

Distance that can be covered by a full tank.

Solution:

To find the distance that will be covered by a full tank, we simply find the product of the rate of fuel usage by the car and tank volume.

Distance covered by full tank = 22mile/gallon x 14gallon = 308miles

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Consider the reaction of A(g) + B(g) + C(g) => D(g) for which the following data were obtained:

Drupady [299]

Answer:

$r = k [A]^{2}[B]^{2}$

Explanation:

A + B + C ⟶ D

$\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}$

Our problem is to determine the values of m, n, and o.

We use the method of initial rates to determine the order of reaction with respect to a component.

(a) Order with respect to A

We must find a pair of experiments in which [A] changes, but [B] and C do not.

They would be Experiments 1 and 2.

[B] and [C] are constant, so only [A] is changing.

$\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}$

(b) Order with respect to B

We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.

They would be Experiments 2 and 3.

[A] and [C] are constant, so only [B] is changing.

$\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}$

(c) Order with respect to C

We must find a pair of experiments in which [C] changes, but [A] and [B] do not.

They would be Experiments 1 and 4.

[A] and [B] are constant, so only [C] is changing.

$\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}$

5 0

3 years ago

What is the molarity of 122.5 g of AlCl3 in 1.0 L of solution? (MM = 133 g/mol)

yawa3891 [41]

Answer: Molarity is defined as moles of solute per liter of solution. So, find the moles of solute and divide by the liters of solution.
molar mass AlCl3 = 133g/mole
moles AlCl3 = 127 g x 1 mole/133 g = 0.955 moles
liters of solution = 400 ml x 1 liter/1000 ml = 0.400 liters
Molarity = 0.955 moles/0.400 liters = 2.39 M
Explain: I looked it up on wyzant.com

4 0

3 years ago

Heating chemicals is defined by the process of combustion.<br> A. True<br> B. False

padilas [110]

Answer: A. True

Explanation: Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives off heat. The original substance is called the fuel, and the source of oxygen is called the oxidizer. ... During combustion, new chemical substances are created from the fuel and the oxidizer.

3 0

3 years ago

A sample of octane undergoes combustion according to the equation 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH°rxn = -11018 kJ. What mas

Anna71 [15]

Answer:

$\large \boxed{\text{528.7 g} }$

Explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be 11018 "moles" of "kJ"

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r: 32.00

2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ

n/mol: 7280

1. Moles of O₂

The molar ratio is 25 mol O₂:11 018 kJ

$\text{Moles of O}_{2} = \text{7280 kJ} \times \dfrac{\text{25 mol O}_{2}}{\text{11 018 kJ}} = \text{16.52 mol O}_{2}$

2. Mass of O₂

$\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}$

3 0

3 years ago

How many moles of CO are present in 35.88 L of the gas?

Pepsi [2]

Answer:

1.602 moles CO

Explanation:

To convert from liters to moles, divide by 22.4:

35.88 L / 22.4 = 1.602 moles CO

7 0

3 years ago